{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0047",
  "pilot_number": "P250-0097",
  "number": 97,
  "family_id": "qf-mod40-binomial-expansion",
  "title": {
    "zh-Hant": "第 97 題：二項式定理",
    "en": "Question 97: 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-0047",
    "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\":21}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], complete_k_zero_through_n_coefficients)",
      "explanation_zh": "Wave6第三層核對=(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], 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": 202611095047,
      "parameters": {
        "n": 21,
        "domain": "Exact_commutative_binomial_theorem",
        "operation": "binomial-expansion",
        "semantic_template": "qf-mod40-binomial-expansion/{\"n\":21}"
      }
    },
    "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-0047",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "38fcaa28ce86999a50dea4366babdfeee8010274dc252d7b8c10a72b2cb04b78",
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095047,
      "semantic_template": "qf-mod40-binomial-expansion/{\"n\":21}"
    },
    "fields": {
      "title": "The Finite Expansion Includes Every k from Zero through n",
      "stem": "Use natural exponent n=21, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).",
      "answer": {
        "kind": "expression",
        "value": "(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], 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 (21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], 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\":21}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use natural exponent n=21, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "n=21≥0; x,y commute",
      "after": "(x+y)^21 is a finite product of 21 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 21 factors to contribute y",
      "after": "C(21,0)=51090942171709440000/(1·51090942171709440000)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 0 個 y 位置與 21 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 0 y positions and 21 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": [
          "21",
          "0"
        ],
        "output": "1",
        "label": "C(21,0)"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C(21,0)=1; choose y 0 times",
      "after": "1·x^21·y^0",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-0=21；指數相加仍為 21。",
        "en": "The x exponent is 21-0=21; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,1)=51090942171709440000/(1·2432902008176640000)=21",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 1 個 y 位置與 20 個 x 位置各自的重排，得到係數 21。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 1 y positions and 20 x positions, giving coefficient 21."
      },
      "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": [
          "21",
          "1"
        ],
        "output": "21",
        "label": "C(21,1)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "C(21,1)=21; choose y 1 times",
      "after": "21·x^20·y^1",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-1=20；指數相加仍為 21。",
        "en": "The x exponent is 21-1=20; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,2)=51090942171709440000/(2·121645100408832000)=210",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 2 個 y 位置與 19 個 x 位置各自的重排，得到係數 210。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 2 y positions and 19 x positions, giving coefficient 210."
      },
      "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": [
          "21",
          "2"
        ],
        "output": "210",
        "label": "C(21,2)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "C(21,2)=210; choose y 2 times",
      "after": "210·x^19·y^2",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-2=19；指數相加仍為 21。",
        "en": "The x exponent is 21-2=19; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,3)=51090942171709440000/(6·6402373705728000)=1330",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 3 個 y 位置與 18 個 x 位置各自的重排，得到係數 1330。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 3 y positions and 18 x positions, giving coefficient 1330."
      },
      "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": [
          "21",
          "3"
        ],
        "output": "1330",
        "label": "C(21,3)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "C(21,3)=1330; choose y 3 times",
      "after": "1330·x^18·y^3",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-3=18；指數相加仍為 21。",
        "en": "The x exponent is 21-3=18; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,4)=51090942171709440000/(24·355687428096000)=5985",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 4 個 y 位置與 17 個 x 位置各自的重排，得到係數 5985。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 4 y positions and 17 x positions, giving coefficient 5985."
      },
      "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": [
          "21",
          "4"
        ],
        "output": "5985",
        "label": "C(21,4)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "C(21,4)=5985; choose y 4 times",
      "after": "5985·x^17·y^4",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-4=17；指數相加仍為 21。",
        "en": "The x exponent is 21-4=17; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,5)=51090942171709440000/(120·20922789888000)=20349",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 5 個 y 位置與 16 個 x 位置各自的重排，得到係數 20349。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 5 y positions and 16 x positions, giving coefficient 20349."
      },
      "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": [
          "21",
          "5"
        ],
        "output": "20349",
        "label": "C(21,5)"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "C(21,5)=20349; choose y 5 times",
      "after": "20349·x^16·y^5",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-5=16；指數相加仍為 21。",
        "en": "The x exponent is 21-5=16; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,6)=51090942171709440000/(720·1307674368000)=54264",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 6 個 y 位置與 15 個 x 位置各自的重排，得到係數 54264。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 6 y positions and 15 x positions, giving coefficient 54264."
      },
      "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": [
          "21",
          "6"
        ],
        "output": "54264",
        "label": "C(21,6)"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "C(21,6)=54264; choose y 6 times",
      "after": "54264·x^15·y^6",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-6=15；指數相加仍為 21。",
        "en": "The x exponent is 21-6=15; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,7)=51090942171709440000/(5040·87178291200)=116280",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 7 個 y 位置與 14 個 x 位置各自的重排，得到係數 116280。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 7 y positions and 14 x positions, giving coefficient 116280."
      },
      "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": [
          "21",
          "7"
        ],
        "output": "116280",
        "label": "C(21,7)"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "C(21,7)=116280; choose y 7 times",
      "after": "116280·x^14·y^7",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-7=14；指數相加仍為 21。",
        "en": "The x exponent is 21-7=14; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,8)=51090942171709440000/(40320·6227020800)=203490",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 8 個 y 位置與 13 個 x 位置各自的重排，得到係數 203490。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 8 y positions and 13 x positions, giving coefficient 203490."
      },
      "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": [
          "21",
          "8"
        ],
        "output": "203490",
        "label": "C(21,8)"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "C(21,8)=203490; choose y 8 times",
      "after": "203490·x^13·y^8",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-8=13；指數相加仍為 21。",
        "en": "The x exponent is 21-8=13; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,9)=51090942171709440000/(362880·479001600)=293930",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 9 個 y 位置與 12 個 x 位置各自的重排，得到係數 293930。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 9 y positions and 12 x positions, giving coefficient 293930."
      },
      "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": [
          "21",
          "9"
        ],
        "output": "293930",
        "label": "C(21,9)"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "C(21,9)=293930; choose y 9 times",
      "after": "293930·x^12·y^9",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-9=12；指數相加仍為 21。",
        "en": "The x exponent is 21-9=12; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,10)=51090942171709440000/(3628800·39916800)=352716",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 10 個 y 位置與 11 個 x 位置各自的重排，得到係數 352716。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 10 y positions and 11 x positions, giving coefficient 352716."
      },
      "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": [
          "21",
          "10"
        ],
        "output": "352716",
        "label": "C(21,10)"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "C(21,10)=352716; choose y 10 times",
      "after": "352716·x^11·y^10",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-10=11；指數相加仍為 21。",
        "en": "The x exponent is 21-10=11; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,11)=51090942171709440000/(39916800·3628800)=352716",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 11 個 y 位置與 10 個 x 位置各自的重排，得到係數 352716。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 11 y positions and 10 x positions, giving coefficient 352716."
      },
      "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": [
          "21",
          "11"
        ],
        "output": "352716",
        "label": "C(21,11)"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "C(21,11)=352716; choose y 11 times",
      "after": "352716·x^10·y^11",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-11=10；指數相加仍為 21。",
        "en": "The x exponent is 21-11=10; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,12)=51090942171709440000/(479001600·362880)=293930",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 12 個 y 位置與 9 個 x 位置各自的重排，得到係數 293930。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 12 y positions and 9 x positions, giving coefficient 293930."
      },
      "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": [
          "21",
          "12"
        ],
        "output": "293930",
        "label": "C(21,12)"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "C(21,12)=293930; choose y 12 times",
      "after": "293930·x^9·y^12",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-12=9；指數相加仍為 21。",
        "en": "The x exponent is 21-12=9; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,13)=51090942171709440000/(6227020800·40320)=203490",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 13 個 y 位置與 8 個 x 位置各自的重排，得到係數 203490。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 13 y positions and 8 x positions, giving coefficient 203490."
      },
      "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": [
          "21",
          "13"
        ],
        "output": "203490",
        "label": "C(21,13)"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "C(21,13)=203490; choose y 13 times",
      "after": "203490·x^8·y^13",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-13=8；指數相加仍為 21。",
        "en": "The x exponent is 21-13=8; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,14)=51090942171709440000/(87178291200·5040)=116280",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 14 個 y 位置與 7 個 x 位置各自的重排，得到係數 116280。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 14 y positions and 7 x positions, giving coefficient 116280."
      },
      "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": [
          "21",
          "14"
        ],
        "output": "116280",
        "label": "C(21,14)"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "C(21,14)=116280; choose y 14 times",
      "after": "116280·x^7·y^14",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-14=7；指數相加仍為 21。",
        "en": "The x exponent is 21-14=7; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,15)=51090942171709440000/(1307674368000·720)=54264",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 15 個 y 位置與 6 個 x 位置各自的重排，得到係數 54264。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 15 y positions and 6 x positions, giving coefficient 54264."
      },
      "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": [
          "21",
          "15"
        ],
        "output": "54264",
        "label": "C(21,15)"
      }
    },
    {
      "id": "S033",
      "uses": [
        "G001",
        "S032"
      ],
      "before": "C(21,15)=54264; choose y 15 times",
      "after": "54264·x^6·y^15",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-15=6；指數相加仍為 21。",
        "en": "The x exponent is 21-15=6; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,16)=51090942171709440000/(20922789888000·120)=20349",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 16 個 y 位置與 5 個 x 位置各自的重排，得到係數 20349。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 16 y positions and 5 x positions, giving coefficient 20349."
      },
      "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": [
          "21",
          "16"
        ],
        "output": "20349",
        "label": "C(21,16)"
      }
    },
    {
      "id": "S035",
      "uses": [
        "G001",
        "S034"
      ],
      "before": "C(21,16)=20349; choose y 16 times",
      "after": "20349·x^5·y^16",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-16=5；指數相加仍為 21。",
        "en": "The x exponent is 21-16=5; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,17)=51090942171709440000/(355687428096000·24)=5985",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 17 個 y 位置與 4 個 x 位置各自的重排，得到係數 5985。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 17 y positions and 4 x positions, giving coefficient 5985."
      },
      "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": [
          "21",
          "17"
        ],
        "output": "5985",
        "label": "C(21,17)"
      }
    },
    {
      "id": "S037",
      "uses": [
        "G001",
        "S036"
      ],
      "before": "C(21,17)=5985; choose y 17 times",
      "after": "5985·x^4·y^17",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-17=4；指數相加仍為 21。",
        "en": "The x exponent is 21-17=4; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,18)=51090942171709440000/(6402373705728000·6)=1330",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 18 個 y 位置與 3 個 x 位置各自的重排，得到係數 1330。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 18 y positions and 3 x positions, giving coefficient 1330."
      },
      "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": [
          "21",
          "18"
        ],
        "output": "1330",
        "label": "C(21,18)"
      }
    },
    {
      "id": "S039",
      "uses": [
        "G001",
        "S038"
      ],
      "before": "C(21,18)=1330; choose y 18 times",
      "after": "1330·x^3·y^18",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-18=3；指數相加仍為 21。",
        "en": "The x exponent is 21-18=3; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,19)=51090942171709440000/(121645100408832000·2)=210",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 19 個 y 位置與 2 個 x 位置各自的重排，得到係數 210。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 19 y positions and 2 x positions, giving coefficient 210."
      },
      "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": [
          "21",
          "19"
        ],
        "output": "210",
        "label": "C(21,19)"
      }
    },
    {
      "id": "S041",
      "uses": [
        "G001",
        "S040"
      ],
      "before": "C(21,19)=210; choose y 19 times",
      "after": "210·x^2·y^19",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-19=2；指數相加仍為 21。",
        "en": "The x exponent is 21-19=2; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,20)=51090942171709440000/(2432902008176640000·1)=21",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 20 個 y 位置與 1 個 x 位置各自的重排，得到係數 21。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 20 y positions and 1 x positions, giving coefficient 21."
      },
      "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": [
          "21",
          "20"
        ],
        "output": "21",
        "label": "C(21,20)"
      }
    },
    {
      "id": "S043",
      "uses": [
        "G001",
        "S042"
      ],
      "before": "C(21,20)=21; choose y 20 times",
      "after": "21·x^1·y^20",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-20=1；指數相加仍為 21。",
        "en": "The x exponent is 21-20=1; the two exponents still sum to 21."
      },
      "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 21 factors to contribute y",
      "after": "C(21,21)=51090942171709440000/(51090942171709440000·1)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 21 個位置，再除掉 21 個 y 位置與 0 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 21 positions and divide by reorderings within the 21 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": [
          "21",
          "21"
        ],
        "output": "1",
        "label": "C(21,21)"
      }
    },
    {
      "id": "S045",
      "uses": [
        "G001",
        "S044"
      ],
      "before": "C(21,21)=1; choose y 21 times",
      "after": "1·x^0·y^21",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 21-21=0；指數相加仍為 21。",
        "en": "The x exponent is 21-21=0; the two exponents still sum to 21."
      },
      "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": "all k=0 through 21 have been accounted for",
      "after": "(x+y)^21=1·x^21·y^0 + 21·x^20·y^1 + 210·x^19·y^2 + 1330·x^18·y^3 + 5985·x^17·y^4 + 20349·x^16·y^5 + 54264·x^15·y^6 + 116280·x^14·y^7 + 203490·x^13·y^8 + 293930·x^12·y^9 + 352716·x^11·y^10 + 352716·x^10·y^11 + 293930·x^9·y^12 + 203490·x^8·y^13 + 116280·x^7·y^14 + 54264·x^6·y^15 + 20349·x^5·y^16 + 5985·x^4·y^17 + 1330·x^3·y^18 + 210·x^2·y^19 + 21·x^1·y^20 + 1·x^0·y^21",
      "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": 21,
    "terms": [
      [
        0,
        1,
        21,
        0
      ],
      [
        1,
        21,
        20,
        1
      ],
      [
        2,
        210,
        19,
        2
      ],
      [
        3,
        1330,
        18,
        3
      ],
      [
        4,
        5985,
        17,
        4
      ],
      [
        5,
        20349,
        16,
        5
      ],
      [
        6,
        54264,
        15,
        6
      ],
      [
        7,
        116280,
        14,
        7
      ],
      [
        8,
        203490,
        13,
        8
      ],
      [
        9,
        293930,
        12,
        9
      ],
      [
        10,
        352716,
        11,
        10
      ],
      [
        11,
        352716,
        10,
        11
      ],
      [
        12,
        293930,
        9,
        12
      ],
      [
        13,
        203490,
        8,
        13
      ],
      [
        14,
        116280,
        7,
        14
      ],
      [
        15,
        54264,
        6,
        15
      ],
      [
        16,
        20349,
        5,
        16
      ],
      [
        17,
        5985,
        4,
        17
      ],
      [
        18,
        1330,
        3,
        18
      ],
      [
        19,
        210,
        2,
        19
      ],
      [
        20,
        21,
        1,
        20
      ],
      [
        21,
        1,
        0,
        21
      ]
    ],
    "boundary": "complete_k_zero_through_n_coefficients"
  },
  "conclusion": {
    "canonical_value": "(21, [[0, 1, 21, 0], [1, 21, 20, 1], [2, 210, 19, 2], [3, 1330, 18, 3], [4, 5985, 17, 4], [5, 20349, 16, 5], [6, 54264, 15, 6], [7, 116280, 14, 7], [8, 203490, 13, 8], [9, 293930, 12, 9], [10, 352716, 11, 10], [11, 352716, 10, 11], [12, 293930, 9, 12], [13, 203490, 8, 13], [14, 116280, 7, 14], [15, 54264, 6, 15], [16, 20349, 5, 16], [17, 5985, 4, 17], [18, 1330, 3, 18], [19, 210, 2, 19], [20, 21, 1, 20], [21, 1, 0, 21]], 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": 47,
      "original_line_sha256": "956ec893f39f7988d3f03b0850c16447aa920803ccffbd66e18f9500743b5c01",
      "question_compact_sha256": "38fcaa28ce86999a50dea4366babdfeee8010274dc252d7b8c10a72b2cb04b78",
      "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": "672502e7e7cf1dd8ea9a2987ba53990b83d4eda01a73bc7b3baea7680341701f"
    },
    "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": "2be0c3d7aa548fde169c606b494e2f05d26e5bd4657bce042befab2e69984bd6",
    "original_candidate_line_sha256": "2be0c3d7aa548fde169c606b494e2f05d26e5bd4657bce042befab2e69984bd6"
  },
  "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": 46,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "33d6899f3c285080c23bb4839f095bf4a53d37c6ac802f8a2f898107623783dd",
      "arithmetic_steps": 22
    },
    "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."
  }
}