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{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0043","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[2, 5]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(10, [2, 10])","explanation_zh":"完整直積與前綴數=(10, [2, 10])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122043,"parameters":{"sizes":[2,5],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[2,5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0044","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[2, 6]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(12, [2, 12])","explanation_zh":"完整直積與前綴數=(12, [2, 12])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122044,"parameters":{"sizes":[2,6],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[2,6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0045","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[2, 7]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(14, [2, 14])","explanation_zh":"完整直積與前綴數=(14, [2, 14])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122045,"parameters":{"sizes":[2,7],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[2,7]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0046","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[2, 8]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(16, [2, 16])","explanation_zh":"完整直積與前綴數=(16, [2, 16])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122046,"parameters":{"sizes":[2,8],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[2,8]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0047","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[3, 1]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(3, [3, 3])","explanation_zh":"完整直積與前綴數=(3, [3, 3])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122047,"parameters":{"sizes":[3,1],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[3,1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0048","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[3, 2]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(6, [3, 6])","explanation_zh":"完整直積與前綴數=(6, [3, 6])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122048,"parameters":{"sizes":[3,2],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[3,2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0049","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[3, 3]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(9, [3, 9])","explanation_zh":"完整直積與前綴數=(9, [3, 9])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122049,"parameters":{"sizes":[3,3],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[3,3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0050","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"完整直積的總數與有序前綴","stem_zh":"有限選擇協議；完整直積階段的選項數依序sizes=[3, 4]，所有跨階段組合均允許。依序答 (直積總數, 各前綴直積數列表)，零階段空直積有一個空選擇序列。","answer":{"kind":"expression","value":"(12, [3, 12])","explanation_zh":"完整直積與前綴數=(12, [3, 12])。逐階段乘入選項數；零階段乘積一，遇零後的前綴皆零，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["完整直積的總數與有序前綴"]},"variant":{"family_id":"qf-density12-counting-full-product-prefix","seed":202610122050,"parameters":{"sizes":[3,4],"operation":"density12_counting_full_product_prefix","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-full-product-prefix/[3,4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0051","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[]，n=0為分支數，待核共同寬度h=7。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(0, 0, 是, 0)","explanation_zh":"條件分支資格與缺陷=(0, 0, 是, 0)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122051,"parameters":{"counts":[],"h":7,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[\"empty_first_stage\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0052","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[1, 3]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(4, 4, 否, 0)","explanation_zh":"條件分支資格與缺陷=(4, 4, 否, 0)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122052,"parameters":{"counts":[1,3],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,4,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0053","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[2, 2]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(4, 4, 是, 0)","explanation_zh":"條件分支資格與缺陷=(4, 4, 是, 0)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122053,"parameters":{"counts":[2,2],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,4,2,true]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0054","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[0]，n=1為分支數，待核共同寬度h=0。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(0, 0, 是, 0)","explanation_zh":"條件分支資格與缺陷=(0, 0, 是, 0)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122054,"parameters":{"counts":[0],"h":0,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[1,0,0,true]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0055","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[3]，n=1為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(3, 2, 否, -1)","explanation_zh":"條件分支資格與缺陷=(3, 2, 否, -1)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122055,"parameters":{"counts":[3],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[1,3,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0056","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[5, 6]，n=2為分支數，待核共同寬度h=0。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(11, 0, 否, -11)","explanation_zh":"條件分支資格與缺陷=(11, 0, 否, -11)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122056,"parameters":{"counts":[5,6],"h":0,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,11,0,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0057","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[6, 7]，n=2為分支數，待核共同寬度h=0。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(13, 0, 否, -13)","explanation_zh":"條件分支資格與缺陷=(13, 0, 否, -13)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122057,"parameters":{"counts":[6,7],"h":0,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,13,0,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0058","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[7, 8]，n=2為分支數，待核共同寬度h=0。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(15, 0, 否, -15)","explanation_zh":"條件分支資格與缺陷=(15, 0, 否, -15)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122058,"parameters":{"counts":[7,8],"h":0,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,15,0,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0059","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[8, 0]，n=2為分支數，待核共同寬度h=0。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(8, 0, 否, -8)","explanation_zh":"條件分支資格與缺陷=(8, 0, 否, -8)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122059,"parameters":{"counts":[8,0],"h":0,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,8,0,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0060","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[0, 1]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(1, 2, 否, 1)","explanation_zh":"條件分支資格與缺陷=(1, 2, 否, 1)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122060,"parameters":{"counts":[0,1],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,1,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0061","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[1, 2]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(3, 2, 否, -1)","explanation_zh":"條件分支資格與缺陷=(3, 2, 否, -1)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122061,"parameters":{"counts":[1,2],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,3,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0062","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[2, 3]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(5, 2, 否, -3)","explanation_zh":"條件分支資格與缺陷=(5, 2, 否, -3)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122062,"parameters":{"counts":[2,3],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,5,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0063","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[3, 4]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(7, 2, 否, -5)","explanation_zh":"條件分支資格與缺陷=(7, 2, 否, -5)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122063,"parameters":{"counts":[3,4],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,7,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0064","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[4, 5]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(9, 2, 否, -7)","explanation_zh":"條件分支資格與缺陷=(9, 2, 否, -7)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122064,"parameters":{"counts":[4,5],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,9,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0065","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[5, 6]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(11, 2, 否, -9)","explanation_zh":"條件分支資格與缺陷=(11, 2, 否, -9)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122065,"parameters":{"counts":[5,6],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,11,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0066","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[6, 7]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(13, 2, 否, -11)","explanation_zh":"條件分支資格與缺陷=(13, 2, 否, -11)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122066,"parameters":{"counts":[6,7],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,13,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0067","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[7, 8]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(15, 2, 否, -13)","explanation_zh":"條件分支資格與缺陷=(15, 2, 否, -13)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122067,"parameters":{"counts":[7,8],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,15,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0068","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[8, 0]，n=2為分支數，待核共同寬度h=1。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(8, 2, 否, -6)","explanation_zh":"條件分支資格與缺陷=(8, 2, 否, -6)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122068,"parameters":{"counts":[8,0],"h":1,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,8,1,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0069","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[0, 1]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(1, 4, 否, 3)","explanation_zh":"條件分支資格與缺陷=(1, 4, 否, 3)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122069,"parameters":{"counts":[0,1],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,1,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0070","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[1, 2]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(3, 4, 否, 1)","explanation_zh":"條件分支資格與缺陷=(3, 4, 否, 1)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122070,"parameters":{"counts":[1,2],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,3,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0071","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[2, 3]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(5, 4, 否, -1)","explanation_zh":"條件分支資格與缺陷=(5, 4, 否, -1)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122071,"parameters":{"counts":[2,3],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,5,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0072","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[3, 4]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(7, 4, 否, -3)","explanation_zh":"條件分支資格與缺陷=(7, 4, 否, -3)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122072,"parameters":{"counts":[3,4],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,7,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0073","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[4, 5]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(9, 4, 否, -5)","explanation_zh":"條件分支資格與缺陷=(9, 4, 否, -5)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122073,"parameters":{"counts":[4,5],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,9,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0074","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[5, 6]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(11, 4, 否, -7)","explanation_zh":"條件分支資格與缺陷=(11, 4, 否, -7)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122074,"parameters":{"counts":[5,6],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,11,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0075","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"條件分支的等寬資格與計數缺陷","stem_zh":"有限選擇協議；第一階段各選擇的第二階段數依序counts=[6, 7]，n=2為分支數，待核共同寬度h=2。依序答 (實際總數, 假定n·h總數, 每個分支是否都等於h, n·h−實際總數)。","answer":{"kind":"expression","value":"(13, 4, 否, -9)","explanation_zh":"條件分支資格與缺陷=(13, 4, 否, -9)。先加實際分支並逐一核對寬度；總數相同不保證等寬，空第一階段條件全稱為是，非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["條件分支的等寬資格與計數缺陷"]},"variant":{"family_id":"qf-density12-counting-conditional-candidate-defect","seed":202610122075,"parameters":{"counts":[6,7],"h":2,"operation":"density12_counting_conditional_candidate_defect","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-conditional-candidate-defect/[2,13,2,false]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0076","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[]、B=[]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(0, 0, 0)","explanation_zh":"重疊前後計數=(0, 0, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122076,"parameters":{"A":[],"B":[],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0077","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0]、B=[0]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(2, 1, 1)","explanation_zh":"重疊前後計數=(2, 1, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122077,"parameters":{"A":[0],"B":[0],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0078","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[]、B=[100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(1, 1, 0)","explanation_zh":"重疊前後計數=(1, 1, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122078,"parameters":{"A":[],"B":[100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,1],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0079","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1]、B=[0, 1]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 2, 2)","explanation_zh":"重疊前後計數=(4, 2, 2)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122079,"parameters":{"A":[0,1],"B":[0,1],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0080","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0]、B=[0, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(3, 2, 1)","explanation_zh":"重疊前後計數=(3, 2, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122080,"parameters":{"A":[0],"B":[0,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,1],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0081","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[]、B=[100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(2, 2, 0)","explanation_zh":"重疊前後計數=(2, 2, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122081,"parameters":{"A":[],"B":[100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,2],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0082","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[-1]、B=[100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(2, 2, 0)","explanation_zh":"重疊前後計數=(2, 2, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122082,"parameters":{"A":[-1],"B":[100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,1],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0083","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2]、B=[0, 1, 2]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(6, 3, 3)","explanation_zh":"重疊前後計數=(6, 3, 3)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":1,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122083,"parameters":{"A":[0,1,2],"B":[0,1,2],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0084","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1]、B=[0, 1, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(5, 3, 2)","explanation_zh":"重疊前後計數=(5, 3, 2)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122084,"parameters":{"A":[0,1],"B":[0,1,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,1],2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0085","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0]、B=[0, 100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 3, 1)","explanation_zh":"重疊前後計數=(4, 3, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122085,"parameters":{"A":[0],"B":[0,100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,2],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0086","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[]、B=[100, 101, 102]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(3, 3, 0)","explanation_zh":"重疊前後計數=(3, 3, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122086,"parameters":{"A":[],"B":[100,101,102],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,3],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0087","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, -1]、B=[0, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 3, 1)","explanation_zh":"重疊前後計數=(4, 3, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122087,"parameters":{"A":[0,-1],"B":[0,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,1],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0088","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[-1]、B=[100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(3, 3, 0)","explanation_zh":"重疊前後計數=(3, 3, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122088,"parameters":{"A":[-1],"B":[100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,2],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0089","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2, 3]、B=[0, 1, 2, 3]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(8, 4, 4)","explanation_zh":"重疊前後計數=(8, 4, 4)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122089,"parameters":{"A":[0,1,2,3],"B":[0,1,2,3],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0090","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2]、B=[0, 1, 2, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(7, 4, 3)","explanation_zh":"重疊前後計數=(7, 4, 3)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122090,"parameters":{"A":[0,1,2],"B":[0,1,2,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,1],3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0091","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1]、B=[0, 1, 100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(6, 4, 2)","explanation_zh":"重疊前後計數=(6, 4, 2)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122091,"parameters":{"A":[0,1],"B":[0,1,100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,2],2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0092","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0]、B=[0, 100, 101, 102]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(5, 4, 1)","explanation_zh":"重疊前後計數=(5, 4, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":2,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122092,"parameters":{"A":[0],"B":[0,100,101,102],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,3],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0093","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[]、B=[100, 101, 102, 103]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 4, 0)","explanation_zh":"重疊前後計數=(4, 4, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122093,"parameters":{"A":[],"B":[100,101,102,103],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,4],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0094","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, -1]、B=[0, 1, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(6, 4, 2)","explanation_zh":"重疊前後計數=(6, 4, 2)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122094,"parameters":{"A":[0,1,-1],"B":[0,1,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,1],2]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0095","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, -1]、B=[0, 100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(5, 4, 1)","explanation_zh":"重疊前後計數=(5, 4, 1)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122095,"parameters":{"A":[0,-1],"B":[0,100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,2],1]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0096","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[-1]、B=[100, 101, 102]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 4, 0)","explanation_zh":"重疊前後計數=(4, 4, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122096,"parameters":{"A":[-1],"B":[100,101,102],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[1,3],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0097","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[-2, -1]、B=[100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(4, 4, 0)","explanation_zh":"重疊前後計數=(4, 4, 0)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122097,"parameters":{"A":[-2,-1],"B":[100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[2,2],0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0098","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2, 3, 4]、B=[0, 1, 2, 3, 4]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(10, 5, 5)","explanation_zh":"重疊前後計數=(10, 5, 5)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122098,"parameters":{"A":[0,1,2,3,4],"B":[0,1,2,3,4],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,0],5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0099","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2, 3]、B=[0, 1, 2, 3, 100]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(9, 5, 4)","explanation_zh":"重疊前後計數=(9, 5, 4)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122099,"parameters":{"A":[0,1,2,3],"B":[0,1,2,3,100],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,1],4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261008-advance-20261012-r02-counting-principle-0100","batch_id":"batch-20261008-advance-20261012-r02-counting-principle-foundations","language":"zh-Hant","source_mko_ids":["mko-counting-principle"],"title_zh":"有重疊選擇的前後計數","stem_zh":"有限選擇協議；整數選擇集合A=[0, 1, 2]、B=[0, 1, 2, 100, 101]。依序求 (|A|+|B|, |A∪B|, |A∩B|)，不要把重疊元素算兩次。","answer":{"kind":"expression","value":"(8, 5, 3)","explanation_zh":"重疊前後計數=(8, 5, 3)。按實際元素隸屬枚舉聯交，兩集合大小相加會重列交集；非概率獨立假設。"},"difficulty":{"band":"lower_secondary","level":3,"skills":["有重疊選擇的前後計數"]},"variant":{"family_id":"qf-density12-counting-overlap-correction","seed":202610122100,"parameters":{"A":[0,1,2],"B":[0,1,2,100,101],"operation":"density12_counting_overlap_correction","domain":"finite_tagged_counting_protocol","semantic_template":"qf-density12-counting-overlap-correction/[[0,2],3]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-08","contract_version":"ocme-question-factory-v0.1"}}