{
  "$schema": "../../../schemas/mko-v0.4.schema.json",
  "schema_version": "mko-v0.4",
  "id": "mko-permutation-combination",
  "semantic_family_id": "sf-permutation-combination",
  "type": "concept",
  "version": "0.10.0",
  "language": "zh-Hant",
  "titles": {
    "zh-Hant": "排列與組合"
  },
  "summary": {
    "zh-Hant": "由 n 個相異物件不重複取 k 個，有序選取是排列，無序選取是組合。"
  },
  "statement": {
    "zh-Hant": "對自然數 n、k 且 0≤k≤n，由 n 個相異物件無放回選取 k 個的排列數為 P(n,k)=n!/(n-k)!，組合數為 C(n,k)=n!/(k!(n-k)!)；k=0 時兩者均為 1。"
  },
  "formula": {
    "tex": "C(n,k)",
    "mathml": "<math xmlns=\"http://www.w3.org/1998/Math/MathML\"><mrow><mi>C</mi><mo>(</mo><mi>n</mi><mo>,</mo><mi>k</mi><mo>)</mo></mrow></math>",
    "semantic_ast": {
      "type": "function_call",
      "callee": {
        "type": "symbol",
        "name": "C",
        "glyph": "C"
      },
      "arguments": [
        {
          "type": "symbol",
          "name": "n",
          "glyph": "n"
        },
        {
          "type": "symbol",
          "name": "k",
          "glyph": "k"
        }
      ]
    },
    "compiler": {
      "id": "ocme-formula-core",
      "version": "0.4.0",
      "source_sha256": "93a523a1eae952e38e258c654617f540701081652fa7432c87df9fed6694c841"
    }
  },
  "assumptions": [
    {
      "id": "a1",
      "zh-Hant": "物件彼此相異，無放回且不可重複，n、k 是自然數並满足 0≤k≤n。"
    },
    {
      "id": "a2",
      "zh-Hant": "排列區分順序；組合忽略順序，空選取計為一種。"
    }
  ],
  "symbols": [
    {
      "token": "C",
      "role": "combination_count",
      "role_zh": "無序不重複選取數",
      "scope": "representative_formula"
    },
    {
      "token": "n",
      "role": "population_size",
      "role_zh": "相異物件總數",
      "scope": "representative_formula"
    },
    {
      "token": "k",
      "role": "selection_size",
      "role_zh": "選取數",
      "scope": "representative_formula"
    }
  ],
  "dependencies": [
    {
      "id": "mko-counting-principle",
      "title_zh": "基本計數原理",
      "reason_zh": "逐次可選數與有序／無序的責任需要計數原理。"
    },
    {
      "id": "mko-factorial",
      "title_zh": "階乘",
      "reason_zh": "公式以自然數階乘表示，包含 0!=1。"
    }
  ],
  "explanation": {
    "paragraphs_zh": [
      "排列依序選擇，每次可選數依次為 n、n-1、…，直到取完 k 個。空排列有一種。",
      "每個 k 元子集合恰有 k! 種排序，所以把排列計數除以 k! 得到組合計數；此關係依物件相異且無放回。",
      "重複選取、不可區分物件、環形排列及額外限制都需要不同計數模型；不把本輪公式無條件套用到它們。"
    ],
    "common_misconception_zh": "沒有先判斷順序、相異性與放回條件，就直接套排列或組合公式。"
  },
  "proofs": [],
  "computational_companions": [],
  "verification": {
    "computational_status": "not_evaluated_no_computational_claim",
    "warning_zh": "BigInt 精確計算和小集合枚舉可檢查指定計數；一般計數關係的形式 Evidence 仍待建立。",
    "evidence_refs": [],
    "producers": [
      {
        "id": "lean-mathlib",
        "adapter": "formal_system",
        "config_path": "formal/lean/lakefile.toml",
        "status": "configured"
      }
    ]
  },
  "formalization": {
    "status": "not_formalized",
    "target_systems": [
      "Lean 4",
      "Mathlib"
    ],
    "next_obligation_zh": "綁定 Finset 子集合、相異有序 k-tuple、Nat.choose 與每個子集合 k! 種排序的證明。"
  },
  "provenance": {
    "origin": "ocme_local_ai_materialization_candidate",
    "note_zh": "依 research/atlas-permutation-combination.md 整理；2026-09-30 脈衝 R06，英文及 Lean 依使用者指示延後。",
    "content_license": "CC-BY-SA-4.0",
    "code_license": "MIT",
    "created_at": "2026-09-30"
  }
}