{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0048",
  "pilot_number": "P250-0098",
  "number": 98,
  "family_id": "qf-mod40-binomial-expansion",
  "title": {
    "zh-Hant": "第 98 題：二項式定理",
    "en": "Question 98: The Finite Expansion Includes Every k from Zero through n"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261109-r05-binomial-theorem-0048",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-binomial-theorem"
    ],
    "title_zh": "二項式定理",
    "stem_zh": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":22}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)",
      "explanation_zh": "Wave6第三層核對=(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)；依完整明示資料獨立重算；coefficients exact且k、n-k exponents不可錯置；n=0給constant one；不套用negative/noninteger exponent或noncommuting variables；combinatorial witness選k個y factors。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 3,
      "skills": [
        "二項式定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095048,
      "parameters": {
        "n": 22,
        "domain": "Exact_commutative_binomial_theorem",
        "operation": "binomial-expansion",
        "semantic_template": "qf-mod40-binomial-expansion/{\"n\":22}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須主端重算；不構成人工審定、single-element stabilizer、finite-prefix BW、single-cover compactness、zero-mass division、noncommutative binomial、incomplete Möbius factorization、bounded-search nonmembership、Lean或Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0048",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "322148fcbbd558ebcdb0e8488271b3b1160106daff85a692ccf064be64afe5d3",
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095048,
      "semantic_template": "qf-mod40-binomial-expansion/{\"n\":22}"
    },
    "fields": {
      "title": "The Finite Expansion Includes Every k from Zero through n",
      "stem": "Use natural exponent n=22, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).",
      "answer": {
        "kind": "expression",
        "value": "(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)",
        "explanation": "Every row is [k,C(n,k),x exponent n−k,y exponent k]. Retain all n+1 rows and exact integer coefficients. At n=0 the expansion is constant1, including the formal zero-exponent convention. Noncommuting variables require a different theorem. The exact result, retaining the requested order, is (22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)."
      },
      "skills": [
        "The Finite Expansion Includes Every k from Zero through n"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "commutative coefficient setting與n in natural including zero。操作=binomial-expansion；資料={\"n\":22}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use natural exponent n=22, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "n=22≥0; x,y commute",
      "after": "(x+y)^22 is a finite product of 22 factors",
      "relation": "implies",
      "rule_id": "finite_product",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每個因子選 x 或 y，且乘法可交換，才能按 y 的選取次數合併同類項。",
        "en": "Choose x or y from each factor; commutativity permits grouping products by the number of y choices."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "此處不是非整數次方的無窮級數，也不套用到不可交換乘法。",
          "en": "This is not an infinite expansion for a noninteger power or a claim for noncommuting multiplication."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "choose exactly 0 of 22 factors to contribute y",
      "after": "C(22,0)=1124000727777607680000/(1·1124000727777607680000)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 0 個 y 位置與 22 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 0 y positions and 22 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "0"
        ],
        "output": "1",
        "label": "C(22,0)"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C(22,0)=1; choose y 0 times",
      "after": "1·x^22·y^0",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-0=22；指數相加仍為 22。",
        "en": "The x exponent is 22-0=22; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "choose exactly 1 of 22 factors to contribute y",
      "after": "C(22,1)=1124000727777607680000/(1·51090942171709440000)=22",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 1 個 y 位置與 21 個 x 位置各自的重排，得到係數 22。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 1 y positions and 21 x positions, giving coefficient 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "1"
        ],
        "output": "22",
        "label": "C(22,1)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "C(22,1)=22; choose y 1 times",
      "after": "22·x^21·y^1",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-1=21；指數相加仍為 22。",
        "en": "The x exponent is 22-1=21; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "choose exactly 2 of 22 factors to contribute y",
      "after": "C(22,2)=1124000727777607680000/(2·2432902008176640000)=231",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 2 個 y 位置與 20 個 x 位置各自的重排，得到係數 231。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 2 y positions and 20 x positions, giving coefficient 231."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "2"
        ],
        "output": "231",
        "label": "C(22,2)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "C(22,2)=231; choose y 2 times",
      "after": "231·x^20·y^2",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-2=20；指數相加仍為 22。",
        "en": "The x exponent is 22-2=20; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "choose exactly 3 of 22 factors to contribute y",
      "after": "C(22,3)=1124000727777607680000/(6·121645100408832000)=1540",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 3 個 y 位置與 19 個 x 位置各自的重排，得到係數 1540。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 3 y positions and 19 x positions, giving coefficient 1540."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "3"
        ],
        "output": "1540",
        "label": "C(22,3)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "C(22,3)=1540; choose y 3 times",
      "after": "1540·x^19·y^3",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-3=19；指數相加仍為 22。",
        "en": "The x exponent is 22-3=19; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "choose exactly 4 of 22 factors to contribute y",
      "after": "C(22,4)=1124000727777607680000/(24·6402373705728000)=7315",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 4 個 y 位置與 18 個 x 位置各自的重排，得到係數 7315。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 4 y positions and 18 x positions, giving coefficient 7315."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "4"
        ],
        "output": "7315",
        "label": "C(22,4)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "C(22,4)=7315; choose y 4 times",
      "after": "7315·x^18·y^4",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-4=18；指數相加仍為 22。",
        "en": "The x exponent is 22-4=18; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S012",
      "uses": [
        "G001",
        "S011"
      ],
      "before": "choose exactly 5 of 22 factors to contribute y",
      "after": "C(22,5)=1124000727777607680000/(120·355687428096000)=26334",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 5 個 y 位置與 17 個 x 位置各自的重排，得到係數 26334。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 5 y positions and 17 x positions, giving coefficient 26334."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "5"
        ],
        "output": "26334",
        "label": "C(22,5)"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "C(22,5)=26334; choose y 5 times",
      "after": "26334·x^17·y^5",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-5=17；指數相加仍為 22。",
        "en": "The x exponent is 22-5=17; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S014",
      "uses": [
        "G001",
        "S013"
      ],
      "before": "choose exactly 6 of 22 factors to contribute y",
      "after": "C(22,6)=1124000727777607680000/(720·20922789888000)=74613",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 6 個 y 位置與 16 個 x 位置各自的重排，得到係數 74613。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 6 y positions and 16 x positions, giving coefficient 74613."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "6"
        ],
        "output": "74613",
        "label": "C(22,6)"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "C(22,6)=74613; choose y 6 times",
      "after": "74613·x^16·y^6",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-6=16；指數相加仍為 22。",
        "en": "The x exponent is 22-6=16; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S016",
      "uses": [
        "G001",
        "S015"
      ],
      "before": "choose exactly 7 of 22 factors to contribute y",
      "after": "C(22,7)=1124000727777607680000/(5040·1307674368000)=170544",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 7 個 y 位置與 15 個 x 位置各自的重排，得到係數 170544。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 7 y positions and 15 x positions, giving coefficient 170544."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "7"
        ],
        "output": "170544",
        "label": "C(22,7)"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "C(22,7)=170544; choose y 7 times",
      "after": "170544·x^15·y^7",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-7=15；指數相加仍為 22。",
        "en": "The x exponent is 22-7=15; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S018",
      "uses": [
        "G001",
        "S017"
      ],
      "before": "choose exactly 8 of 22 factors to contribute y",
      "after": "C(22,8)=1124000727777607680000/(40320·87178291200)=319770",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 8 個 y 位置與 14 個 x 位置各自的重排，得到係數 319770。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 8 y positions and 14 x positions, giving coefficient 319770."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "8"
        ],
        "output": "319770",
        "label": "C(22,8)"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "C(22,8)=319770; choose y 8 times",
      "after": "319770·x^14·y^8",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-8=14；指數相加仍為 22。",
        "en": "The x exponent is 22-8=14; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S020",
      "uses": [
        "G001",
        "S019"
      ],
      "before": "choose exactly 9 of 22 factors to contribute y",
      "after": "C(22,9)=1124000727777607680000/(362880·6227020800)=497420",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 9 個 y 位置與 13 個 x 位置各自的重排，得到係數 497420。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 9 y positions and 13 x positions, giving coefficient 497420."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "9"
        ],
        "output": "497420",
        "label": "C(22,9)"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "C(22,9)=497420; choose y 9 times",
      "after": "497420·x^13·y^9",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-9=13；指數相加仍為 22。",
        "en": "The x exponent is 22-9=13; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S022",
      "uses": [
        "G001",
        "S021"
      ],
      "before": "choose exactly 10 of 22 factors to contribute y",
      "after": "C(22,10)=1124000727777607680000/(3628800·479001600)=646646",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 10 個 y 位置與 12 個 x 位置各自的重排，得到係數 646646。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 10 y positions and 12 x positions, giving coefficient 646646."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "10"
        ],
        "output": "646646",
        "label": "C(22,10)"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "C(22,10)=646646; choose y 10 times",
      "after": "646646·x^12·y^10",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-10=12；指數相加仍為 22。",
        "en": "The x exponent is 22-10=12; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S024",
      "uses": [
        "G001",
        "S023"
      ],
      "before": "choose exactly 11 of 22 factors to contribute y",
      "after": "C(22,11)=1124000727777607680000/(39916800·39916800)=705432",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 11 個 y 位置與 11 個 x 位置各自的重排，得到係數 705432。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 11 y positions and 11 x positions, giving coefficient 705432."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "11"
        ],
        "output": "705432",
        "label": "C(22,11)"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "C(22,11)=705432; choose y 11 times",
      "after": "705432·x^11·y^11",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-11=11；指數相加仍為 22。",
        "en": "The x exponent is 22-11=11; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S026",
      "uses": [
        "G001",
        "S025"
      ],
      "before": "choose exactly 12 of 22 factors to contribute y",
      "after": "C(22,12)=1124000727777607680000/(479001600·3628800)=646646",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 12 個 y 位置與 10 個 x 位置各自的重排，得到係數 646646。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 12 y positions and 10 x positions, giving coefficient 646646."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "12"
        ],
        "output": "646646",
        "label": "C(22,12)"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "C(22,12)=646646; choose y 12 times",
      "after": "646646·x^10·y^12",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-12=10；指數相加仍為 22。",
        "en": "The x exponent is 22-12=10; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S028",
      "uses": [
        "G001",
        "S027"
      ],
      "before": "choose exactly 13 of 22 factors to contribute y",
      "after": "C(22,13)=1124000727777607680000/(6227020800·362880)=497420",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 13 個 y 位置與 9 個 x 位置各自的重排，得到係數 497420。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 13 y positions and 9 x positions, giving coefficient 497420."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "13"
        ],
        "output": "497420",
        "label": "C(22,13)"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "C(22,13)=497420; choose y 13 times",
      "after": "497420·x^9·y^13",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-13=9；指數相加仍為 22。",
        "en": "The x exponent is 22-13=9; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S030",
      "uses": [
        "G001",
        "S029"
      ],
      "before": "choose exactly 14 of 22 factors to contribute y",
      "after": "C(22,14)=1124000727777607680000/(87178291200·40320)=319770",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 14 個 y 位置與 8 個 x 位置各自的重排，得到係數 319770。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 14 y positions and 8 x positions, giving coefficient 319770."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "14"
        ],
        "output": "319770",
        "label": "C(22,14)"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "C(22,14)=319770; choose y 14 times",
      "after": "319770·x^8·y^14",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-14=8；指數相加仍為 22。",
        "en": "The x exponent is 22-14=8; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S032",
      "uses": [
        "G001",
        "S031"
      ],
      "before": "choose exactly 15 of 22 factors to contribute y",
      "after": "C(22,15)=1124000727777607680000/(1307674368000·5040)=170544",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 15 個 y 位置與 7 個 x 位置各自的重排，得到係數 170544。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 15 y positions and 7 x positions, giving coefficient 170544."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "15"
        ],
        "output": "170544",
        "label": "C(22,15)"
      }
    },
    {
      "id": "S033",
      "uses": [
        "G001",
        "S032"
      ],
      "before": "C(22,15)=170544; choose y 15 times",
      "after": "170544·x^7·y^15",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-15=7；指數相加仍為 22。",
        "en": "The x exponent is 22-15=7; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S034",
      "uses": [
        "G001",
        "S033"
      ],
      "before": "choose exactly 16 of 22 factors to contribute y",
      "after": "C(22,16)=1124000727777607680000/(20922789888000·720)=74613",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 16 個 y 位置與 6 個 x 位置各自的重排，得到係數 74613。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 16 y positions and 6 x positions, giving coefficient 74613."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "16"
        ],
        "output": "74613",
        "label": "C(22,16)"
      }
    },
    {
      "id": "S035",
      "uses": [
        "G001",
        "S034"
      ],
      "before": "C(22,16)=74613; choose y 16 times",
      "after": "74613·x^6·y^16",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-16=6；指數相加仍為 22。",
        "en": "The x exponent is 22-16=6; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S036",
      "uses": [
        "G001",
        "S035"
      ],
      "before": "choose exactly 17 of 22 factors to contribute y",
      "after": "C(22,17)=1124000727777607680000/(355687428096000·120)=26334",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 17 個 y 位置與 5 個 x 位置各自的重排，得到係數 26334。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 17 y positions and 5 x positions, giving coefficient 26334."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "17"
        ],
        "output": "26334",
        "label": "C(22,17)"
      }
    },
    {
      "id": "S037",
      "uses": [
        "G001",
        "S036"
      ],
      "before": "C(22,17)=26334; choose y 17 times",
      "after": "26334·x^5·y^17",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-17=5；指數相加仍為 22。",
        "en": "The x exponent is 22-17=5; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S038",
      "uses": [
        "G001",
        "S037"
      ],
      "before": "choose exactly 18 of 22 factors to contribute y",
      "after": "C(22,18)=1124000727777607680000/(6402373705728000·24)=7315",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 18 個 y 位置與 4 個 x 位置各自的重排，得到係數 7315。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 18 y positions and 4 x positions, giving coefficient 7315."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "18"
        ],
        "output": "7315",
        "label": "C(22,18)"
      }
    },
    {
      "id": "S039",
      "uses": [
        "G001",
        "S038"
      ],
      "before": "C(22,18)=7315; choose y 18 times",
      "after": "7315·x^4·y^18",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-18=4；指數相加仍為 22。",
        "en": "The x exponent is 22-18=4; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S040",
      "uses": [
        "G001",
        "S039"
      ],
      "before": "choose exactly 19 of 22 factors to contribute y",
      "after": "C(22,19)=1124000727777607680000/(121645100408832000·6)=1540",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 19 個 y 位置與 3 個 x 位置各自的重排，得到係數 1540。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 19 y positions and 3 x positions, giving coefficient 1540."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "19"
        ],
        "output": "1540",
        "label": "C(22,19)"
      }
    },
    {
      "id": "S041",
      "uses": [
        "G001",
        "S040"
      ],
      "before": "C(22,19)=1540; choose y 19 times",
      "after": "1540·x^3·y^19",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-19=3；指數相加仍為 22。",
        "en": "The x exponent is 22-19=3; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S042",
      "uses": [
        "G001",
        "S041"
      ],
      "before": "choose exactly 20 of 22 factors to contribute y",
      "after": "C(22,20)=1124000727777607680000/(2432902008176640000·2)=231",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 20 個 y 位置與 2 個 x 位置各自的重排，得到係數 231。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 20 y positions and 2 x positions, giving coefficient 231."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "20"
        ],
        "output": "231",
        "label": "C(22,20)"
      }
    },
    {
      "id": "S043",
      "uses": [
        "G001",
        "S042"
      ],
      "before": "C(22,20)=231; choose y 20 times",
      "after": "231·x^2·y^20",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-20=2；指數相加仍為 22。",
        "en": "The x exponent is 22-20=2; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S044",
      "uses": [
        "G001",
        "S043"
      ],
      "before": "choose exactly 21 of 22 factors to contribute y",
      "after": "C(22,21)=1124000727777607680000/(51090942171709440000·1)=22",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 21 個 y 位置與 1 個 x 位置各自的重排，得到係數 22。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 21 y positions and 1 x positions, giving coefficient 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "21"
        ],
        "output": "22",
        "label": "C(22,21)"
      }
    },
    {
      "id": "S045",
      "uses": [
        "G001",
        "S044"
      ],
      "before": "C(22,21)=22; choose y 21 times",
      "after": "22·x^1·y^21",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-21=1；指數相加仍為 22。",
        "en": "The x exponent is 22-21=1; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S046",
      "uses": [
        "G001",
        "S045"
      ],
      "before": "choose exactly 22 of 22 factors to contribute y",
      "after": "C(22,22)=1124000727777607680000/(1124000727777607680000·1)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 22 個位置，再除掉 22 個 y 位置與 0 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 22 positions and divide by reorderings within the 22 y positions and 0 x positions, giving coefficient 1."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "可用 Pascal 遞迴的獨立整數運算回核係數，不用近似階乘。",
          "en": "Check the coefficient independently with the integer Pascal recurrence, not approximate factorials."
        }
      },
      "arithmetic": {
        "operation": "binomial_count",
        "inputs": [
          "22",
          "22"
        ],
        "output": "1",
        "label": "C(22,22)"
      }
    },
    {
      "id": "S047",
      "uses": [
        "G001",
        "S046"
      ],
      "before": "C(22,22)=1; choose y 22 times",
      "after": "1·x^0·y^22",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 22-22=0；指數相加仍為 22。",
        "en": "The x exponent is 22-22=0; the two exponents still sum to 22."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "原答案的欄序是 [k,係數,x指數,y指數]，不能交換指數欄。",
          "en": "The answer row is [k,coefficient,x exponent,y exponent]; do not swap the exponent columns."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S048",
      "uses": [
        "G001",
        "S047"
      ],
      "before": "all k=0 through 22 have been accounted for",
      "after": "(x+y)^22=1·x^22·y^0 + 22·x^21·y^1 + 231·x^20·y^2 + 1540·x^19·y^3 + 7315·x^18·y^4 + 26334·x^17·y^5 + 74613·x^16·y^6 + 170544·x^15·y^7 + 319770·x^14·y^8 + 497420·x^13·y^9 + 646646·x^12·y^10 + 705432·x^11·y^11 + 646646·x^10·y^12 + 497420·x^9·y^13 + 319770·x^8·y^14 + 170544·x^7·y^15 + 74613·x^6·y^16 + 26334·x^5·y^17 + 7315·x^4·y^18 + 1540·x^3·y^19 + 231·x^2·y^20 + 22·x^1·y^21 + 1·x^0·y^22",
      "relation": "implies",
      "rule_id": "collect_all_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "每一種 x/y 選取都落入唯一的 k 類，因此這是完整有限展開，不是只列頭尾項。",
        "en": "Every x/y choice belongs to exactly one k class, so this is the complete finite expansion, not just endpoint terms."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "共有 n+1 個係數位置，k 的全範圍逐一保留。",
          "en": "There are n+1 coefficient positions, with the complete range of k retained."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "n": 22,
    "terms": [
      [
        0,
        1,
        22,
        0
      ],
      [
        1,
        22,
        21,
        1
      ],
      [
        2,
        231,
        20,
        2
      ],
      [
        3,
        1540,
        19,
        3
      ],
      [
        4,
        7315,
        18,
        4
      ],
      [
        5,
        26334,
        17,
        5
      ],
      [
        6,
        74613,
        16,
        6
      ],
      [
        7,
        170544,
        15,
        7
      ],
      [
        8,
        319770,
        14,
        8
      ],
      [
        9,
        497420,
        13,
        9
      ],
      [
        10,
        646646,
        12,
        10
      ],
      [
        11,
        705432,
        11,
        11
      ],
      [
        12,
        646646,
        10,
        12
      ],
      [
        13,
        497420,
        9,
        13
      ],
      [
        14,
        319770,
        8,
        14
      ],
      [
        15,
        170544,
        7,
        15
      ],
      [
        16,
        74613,
        6,
        16
      ],
      [
        17,
        26334,
        5,
        17
      ],
      [
        18,
        7315,
        4,
        18
      ],
      [
        19,
        1540,
        3,
        19
      ],
      [
        20,
        231,
        2,
        20
      ],
      [
        21,
        22,
        1,
        21
      ],
      [
        22,
        1,
        0,
        22
      ]
    ],
    "boundary": "complete_k_zero_through_n_coefficients"
  },
  "conclusion": {
    "canonical_value": "(22, [[0, 1, 22, 0], [1, 22, 21, 1], [2, 231, 20, 2], [3, 1540, 19, 3], [4, 7315, 18, 4], [5, 26334, 17, 5], [6, 74613, 16, 6], [7, 170544, 15, 7], [8, 319770, 14, 8], [9, 497420, 13, 9], [10, 646646, 12, 10], [11, 705432, 11, 11], [12, 646646, 10, 12], [13, 497420, 9, 13], [14, 319770, 8, 14], [15, 170544, 7, 15], [16, 74613, 6, 16], [17, 26334, 5, 17], [18, 7315, 4, 18], [19, 1540, 3, 19], [20, 231, 2, 20], [21, 22, 1, 21], [22, 1, 0, 22]], complete_k_zero_through_n_coefficients)",
    "explanation": {
      "zh-Hant": "完整展開如上述所有單項式；欄序表逐項對應。",
      "en": "The full expansion consists of all displayed monomials, in the same order as the answer rows."
    }
  },
  "definition_note": {
    "zh-Hant": "n 是非負整數，x 與 y 可交換；n=0 使用多項式空乘積單位元。",
    "en": "n is a nonnegative integer and x,y commute. For n=0 use the polynomial empty-product identity."
  },
  "source": {
    "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db",
    "question_binding": {
      "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
      "batch_path": "public/data/questions/batches/batch-20261014-advance-20261109-r05-binomial-theorem-foundations/questions.jsonl",
      "batch_sha256": "f83302c1353681ecb1209ee1ccbef0d266f15fb9415191ed9289eb329e1175cc",
      "original_line": 48,
      "original_line_sha256": "e22e1783cb61a284cd312d5178d87059f3e547ed0306ff53731d4e07b73330f0",
      "question_compact_sha256": "322148fcbbd558ebcdb0e8488271b3b1160106daff85a692ccf064be64afe5d3",
      "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261109-r05-binomial-theorem-foundations.en.json",
      "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261109-r05-binomial-theorem-foundations.en.jsonl",
      "english_row_compact_sha256": "fde97e91e063d5cef205876b799131c2ca206d238573f1ad49c548b0fe8fe122"
    },
    "mko_bindings": {
      "mko-binomial-theorem": {
        "sha256": "c8fefe3783fc4f7633b4b22b8cf149d96a683900f181e3b42c690409ce90e5b7",
        "english_sha256": "1086dfe7c31f4e6bf8934681a0a080b50c14cb828f1a8b17e59143ec2cf06c83",
        "original_path": "public/data/mko/binomial-theorem.json",
        "english_path": "public/data/i18n/editions/mko-binomial-theorem.en.json"
      }
    },
    "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff",
    "selected_part_sha256": "fb325b42cfcaa25ddbae992459c2684afd767fe218028bc0b9c0823e67c83b7c",
    "original_candidate_sha256": "719658fb3783c22767b4908207071177de5be44e8898f01fffca4e34c07d9242",
    "original_candidate_line_sha256": "719658fb3783c22767b4908207071177de5be44e8898f01fffca4e34c07d9242"
  },
  "historical_source_state": {
    "review_state": {
      "derivation": "candidate_authored_and_locally_checked",
      "source_question_state": "candidate",
      "independent_AI_or_human_review": "NOT_PERFORMED",
      "formalization": "not_formalized",
      "Lean": "NOT_RUN"
    },
    "scope_note": {
      "zh-Hant": "本稿是來源綁定的教學推導候選；精確重算不等於獨立數學審查或形式化證明。原題、答案、公式、AST 與歷史未被修改。",
      "en": "This is a source-bound pedagogical derivation candidate. Exact recomputation is not independent mathematical review or formal proof. Original questions, answers, formulas, ASTs and history are unchanged."
    },
    "label": "Frozen producer metadata; current per-question review is separate below."
  },
  "revision": {
    "id": "review-20261024-002",
    "review_date": "2026-10-24",
    "owner_authorized_reader_publication": true,
    "scope": "Family prototypes and mathematical/bilingual reasoning variants reviewed in this existing context; parameter substitutions, all original/source bindings, full answer tuples and arithmetic ASTs independently reconstructed per question. Numeric binomial text slots are validated against independently regenerated n,k,C(n,k),n-k values. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.",
    "review_receipts": {
      "semantic_review_sha256": "4727d1bd957a9b7f6cadb7972ddb5ab80a31f5259dc5394c1dff2b71944ed36b",
      "question_first_witness_sha256": "51e18649133238ac992229f6e16542a5f3af53ca73b2262ab0524508ea937dee"
    },
    "status": "second_context_reviewed",
    "mathematical_semantics": true,
    "bilingual_semantics": true,
    "steps_reviewed": 48,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "a43251e101277be4fc444fca3ecf18243a3fbc6d69659322b46bffa16b61a628",
      "arithmetic_steps": 23
    },
    "uses_answer_as_premise": false,
    "human_review": "NOT_PERFORMED",
    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}