{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261109-r05-binomial-theorem-0040",
  "pilot_number": "P250-0090",
  "number": 90,
  "family_id": "qf-mod40-binomial-expansion",
  "title": {
    "zh-Hant": "第 90 題：二項式定理",
    "en": "Question 90: 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-0040",
    "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\":14}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
    "answer": {
      "kind": "expression",
      "value": "(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], complete_k_zero_through_n_coefficients)",
      "explanation_zh": "Wave6第三層核對=(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], 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": 2,
      "skills": [
        "二項式定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095040,
      "parameters": {
        "n": 14,
        "domain": "Exact_commutative_binomial_theorem",
        "operation": "binomial-expansion",
        "semantic_template": "qf-mod40-binomial-expansion/{\"n\":14}"
      }
    },
    "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-0040",
    "batch_id": "batch-20261014-advance-20261109-r05-binomial-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "fb267b9df094c9f841f151f97bdc090351cdc90a04eb650e045ed9a392c96c47",
      "family_id": "qf-mod40-binomial-expansion",
      "seed": 202611095040,
      "semantic_template": "qf-mod40-binomial-expansion/{\"n\":14}"
    },
    "fields": {
      "title": "The Finite Expansion Includes Every k from Zero through n",
      "stem": "Use natural exponent n=14, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary).",
      "answer": {
        "kind": "expression",
        "value": "(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], 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 (14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], 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\":14}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。",
        "en": "Use natural exponent n=14, including0, in a commutative coefficient setting. Give (n, coefficient_exponent_rows, boundary)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "n=14≥0; x,y commute",
      "after": "(x+y)^14 is a finite product of 14 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 14 factors to contribute y",
      "after": "C(14,0)=87178291200/(1·87178291200)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 0 個 y 位置與 14 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 0 y positions and 14 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": [
          "14",
          "0"
        ],
        "output": "1",
        "label": "C(14,0)"
      }
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "C(14,0)=1; choose y 0 times",
      "after": "1·x^14·y^0",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-0=14；指數相加仍為 14。",
        "en": "The x exponent is 14-0=14; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,1)=87178291200/(1·6227020800)=14",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 1 個 y 位置與 13 個 x 位置各自的重排，得到係數 14。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 1 y positions and 13 x positions, giving coefficient 14."
      },
      "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": [
          "14",
          "1"
        ],
        "output": "14",
        "label": "C(14,1)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "C(14,1)=14; choose y 1 times",
      "after": "14·x^13·y^1",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-1=13；指數相加仍為 14。",
        "en": "The x exponent is 14-1=13; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,2)=87178291200/(2·479001600)=91",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 2 個 y 位置與 12 個 x 位置各自的重排，得到係數 91。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 2 y positions and 12 x positions, giving coefficient 91."
      },
      "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": [
          "14",
          "2"
        ],
        "output": "91",
        "label": "C(14,2)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "C(14,2)=91; choose y 2 times",
      "after": "91·x^12·y^2",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-2=12；指數相加仍為 14。",
        "en": "The x exponent is 14-2=12; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,3)=87178291200/(6·39916800)=364",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 3 個 y 位置與 11 個 x 位置各自的重排，得到係數 364。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 3 y positions and 11 x positions, giving coefficient 364."
      },
      "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": [
          "14",
          "3"
        ],
        "output": "364",
        "label": "C(14,3)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "C(14,3)=364; choose y 3 times",
      "after": "364·x^11·y^3",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-3=11；指數相加仍為 14。",
        "en": "The x exponent is 14-3=11; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,4)=87178291200/(24·3628800)=1001",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 4 個 y 位置與 10 個 x 位置各自的重排，得到係數 1001。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 4 y positions and 10 x positions, giving coefficient 1001."
      },
      "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": [
          "14",
          "4"
        ],
        "output": "1001",
        "label": "C(14,4)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "C(14,4)=1001; choose y 4 times",
      "after": "1001·x^10·y^4",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-4=10；指數相加仍為 14。",
        "en": "The x exponent is 14-4=10; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,5)=87178291200/(120·362880)=2002",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 5 個 y 位置與 9 個 x 位置各自的重排，得到係數 2002。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 5 y positions and 9 x positions, giving coefficient 2002."
      },
      "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": [
          "14",
          "5"
        ],
        "output": "2002",
        "label": "C(14,5)"
      }
    },
    {
      "id": "S013",
      "uses": [
        "G001",
        "S012"
      ],
      "before": "C(14,5)=2002; choose y 5 times",
      "after": "2002·x^9·y^5",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-5=9；指數相加仍為 14。",
        "en": "The x exponent is 14-5=9; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,6)=87178291200/(720·40320)=3003",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 6 個 y 位置與 8 個 x 位置各自的重排，得到係數 3003。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 6 y positions and 8 x positions, giving coefficient 3003."
      },
      "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": [
          "14",
          "6"
        ],
        "output": "3003",
        "label": "C(14,6)"
      }
    },
    {
      "id": "S015",
      "uses": [
        "G001",
        "S014"
      ],
      "before": "C(14,6)=3003; choose y 6 times",
      "after": "3003·x^8·y^6",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-6=8；指數相加仍為 14。",
        "en": "The x exponent is 14-6=8; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,7)=87178291200/(5040·5040)=3432",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 7 個 y 位置與 7 個 x 位置各自的重排，得到係數 3432。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 7 y positions and 7 x positions, giving coefficient 3432."
      },
      "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": [
          "14",
          "7"
        ],
        "output": "3432",
        "label": "C(14,7)"
      }
    },
    {
      "id": "S017",
      "uses": [
        "G001",
        "S016"
      ],
      "before": "C(14,7)=3432; choose y 7 times",
      "after": "3432·x^7·y^7",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-7=7；指數相加仍為 14。",
        "en": "The x exponent is 14-7=7; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,8)=87178291200/(40320·720)=3003",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 8 個 y 位置與 6 個 x 位置各自的重排，得到係數 3003。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 8 y positions and 6 x positions, giving coefficient 3003."
      },
      "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": [
          "14",
          "8"
        ],
        "output": "3003",
        "label": "C(14,8)"
      }
    },
    {
      "id": "S019",
      "uses": [
        "G001",
        "S018"
      ],
      "before": "C(14,8)=3003; choose y 8 times",
      "after": "3003·x^6·y^8",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-8=6；指數相加仍為 14。",
        "en": "The x exponent is 14-8=6; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,9)=87178291200/(362880·120)=2002",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 9 個 y 位置與 5 個 x 位置各自的重排，得到係數 2002。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 9 y positions and 5 x positions, giving coefficient 2002."
      },
      "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": [
          "14",
          "9"
        ],
        "output": "2002",
        "label": "C(14,9)"
      }
    },
    {
      "id": "S021",
      "uses": [
        "G001",
        "S020"
      ],
      "before": "C(14,9)=2002; choose y 9 times",
      "after": "2002·x^5·y^9",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-9=5；指數相加仍為 14。",
        "en": "The x exponent is 14-9=5; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,10)=87178291200/(3628800·24)=1001",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 10 個 y 位置與 4 個 x 位置各自的重排，得到係數 1001。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 10 y positions and 4 x positions, giving coefficient 1001."
      },
      "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": [
          "14",
          "10"
        ],
        "output": "1001",
        "label": "C(14,10)"
      }
    },
    {
      "id": "S023",
      "uses": [
        "G001",
        "S022"
      ],
      "before": "C(14,10)=1001; choose y 10 times",
      "after": "1001·x^4·y^10",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-10=4；指數相加仍為 14。",
        "en": "The x exponent is 14-10=4; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,11)=87178291200/(39916800·6)=364",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 11 個 y 位置與 3 個 x 位置各自的重排，得到係數 364。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 11 y positions and 3 x positions, giving coefficient 364."
      },
      "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": [
          "14",
          "11"
        ],
        "output": "364",
        "label": "C(14,11)"
      }
    },
    {
      "id": "S025",
      "uses": [
        "G001",
        "S024"
      ],
      "before": "C(14,11)=364; choose y 11 times",
      "after": "364·x^3·y^11",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-11=3；指數相加仍為 14。",
        "en": "The x exponent is 14-11=3; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,12)=87178291200/(479001600·2)=91",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 12 個 y 位置與 2 個 x 位置各自的重排，得到係數 91。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 12 y positions and 2 x positions, giving coefficient 91."
      },
      "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": [
          "14",
          "12"
        ],
        "output": "91",
        "label": "C(14,12)"
      }
    },
    {
      "id": "S027",
      "uses": [
        "G001",
        "S026"
      ],
      "before": "C(14,12)=91; choose y 12 times",
      "after": "91·x^2·y^12",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-12=2；指數相加仍為 14。",
        "en": "The x exponent is 14-12=2; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,13)=87178291200/(6227020800·1)=14",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 13 個 y 位置與 1 個 x 位置各自的重排，得到係數 14。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 13 y positions and 1 x positions, giving coefficient 14."
      },
      "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": [
          "14",
          "13"
        ],
        "output": "14",
        "label": "C(14,13)"
      }
    },
    {
      "id": "S029",
      "uses": [
        "G001",
        "S028"
      ],
      "before": "C(14,13)=14; choose y 13 times",
      "after": "14·x^1·y^13",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-13=1；指數相加仍為 14。",
        "en": "The x exponent is 14-13=1; the two exponents still sum to 14."
      },
      "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 14 factors to contribute y",
      "after": "C(14,14)=87178291200/(87178291200·1)=1",
      "relation": "implies",
      "rule_id": "subset_count",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先排列 14 個位置，再除掉 14 個 y 位置與 0 個 x 位置各自的重排，得到係數 1。",
        "en": "Count orderings of 14 positions and divide by reorderings within the 14 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": [
          "14",
          "14"
        ],
        "output": "1",
        "label": "C(14,14)"
      }
    },
    {
      "id": "S031",
      "uses": [
        "G001",
        "S030"
      ],
      "before": "C(14,14)=1; choose y 14 times",
      "after": "1·x^0·y^14",
      "relation": "implies",
      "rule_id": "group_like_terms",
      "conditions": [],
      "explanation": {
        "zh-Hant": "相應的 x 次數為 14-14=0；指數相加仍為 14。",
        "en": "The x exponent is 14-14=0; the two exponents still sum to 14."
      },
      "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": "all k=0 through 14 have been accounted for",
      "after": "(x+y)^14=1·x^14·y^0 + 14·x^13·y^1 + 91·x^12·y^2 + 364·x^11·y^3 + 1001·x^10·y^4 + 2002·x^9·y^5 + 3003·x^8·y^6 + 3432·x^7·y^7 + 3003·x^6·y^8 + 2002·x^5·y^9 + 1001·x^4·y^10 + 364·x^3·y^11 + 91·x^2·y^12 + 14·x^1·y^13 + 1·x^0·y^14",
      "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": 14,
    "terms": [
      [
        0,
        1,
        14,
        0
      ],
      [
        1,
        14,
        13,
        1
      ],
      [
        2,
        91,
        12,
        2
      ],
      [
        3,
        364,
        11,
        3
      ],
      [
        4,
        1001,
        10,
        4
      ],
      [
        5,
        2002,
        9,
        5
      ],
      [
        6,
        3003,
        8,
        6
      ],
      [
        7,
        3432,
        7,
        7
      ],
      [
        8,
        3003,
        6,
        8
      ],
      [
        9,
        2002,
        5,
        9
      ],
      [
        10,
        1001,
        4,
        10
      ],
      [
        11,
        364,
        3,
        11
      ],
      [
        12,
        91,
        2,
        12
      ],
      [
        13,
        14,
        1,
        13
      ],
      [
        14,
        1,
        0,
        14
      ]
    ],
    "boundary": "complete_k_zero_through_n_coefficients"
  },
  "conclusion": {
    "canonical_value": "(14, [[0, 1, 14, 0], [1, 14, 13, 1], [2, 91, 12, 2], [3, 364, 11, 3], [4, 1001, 10, 4], [5, 2002, 9, 5], [6, 3003, 8, 6], [7, 3432, 7, 7], [8, 3003, 6, 8], [9, 2002, 5, 9], [10, 1001, 4, 10], [11, 364, 3, 11], [12, 91, 2, 12], [13, 14, 1, 13], [14, 1, 0, 14]], 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": 40,
      "original_line_sha256": "654ce81a578f1e36cdc7e083a6bd9a6f552d2f296021b77875b8938dddb2f3f4",
      "question_compact_sha256": "fb267b9df094c9f841f151f97bdc090351cdc90a04eb650e045ed9a392c96c47",
      "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": "2935a9e39086ea16792da7dd71b106d540306510e0a3c97a0bc4e577106aa65e"
    },
    "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": "860c8dd551bb80ff9fa7d7d8f5f308db707d8d16d7b1a4840fe584fca53d702c",
    "original_candidate_line_sha256": "860c8dd551bb80ff9fa7d7d8f5f308db707d8d16d7b1a4840fe584fca53d702c"
  },
  "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": 32,
    "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction",
    "exact_question_first_check": {
      "model_sha256": "444fb02ad1d892f7a995c532abdce109f3abc9d8077b3e816321eea0c72e4f85",
      "arithmetic_steps": 15
    },
    "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."
  }
}