Compute the number of combinations C(n, k) without repetition.
从 n 个元素取 r 个的组合数。
| Item | Formula / Note | Source / Standard |
|---|---|---|
| Combination count | C(n,k) = n! ÷ [k!(n−k)!], unordered selection | Standard combinatorics definition |
| Symmetry identity | C(n,k) = C(n, n−k); use the smaller k to avoid overflow | Consequence of Pascal's rule |
| Pascal recurrence | C(n,k) = C(n−1,k−1) + C(n−1,k), the Yang Hui triangle (c. 1050) | Yang Hui 1261 / Pascal 1654 |
| Combination identities | Σ C(n,k) = 2^n; Vandermonde ΣC(r,k)C(s,n−k) = C(r+s,n) | Standard combinatorial identities |