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从n个不同的元素中，每次取出k个不同的元素，不管其顺序合并成一组，称为组合．其组合种数为 \[ C_n^k = \frac{{A_n^k }}{{k!}} = \frac{{n!}}{{\left( {n - k} \right)!k!}} \] 其中\[ C_n^k 也记作\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) \]
把n个不同的元素分成m组，第i组有ni个不同的元素，即n1+n2+…+nm=n，这样分组的种数为 \[ C_n^{n_1 ,n_2 ,n_3 , \cdots ,n_m } = \frac{{n!}}{{n_1 !n_2 ! \cdots n_m !}} \] 通常意义下的组合是其特例．
从n个不同元素中，每次取出k个元素，允许重复，不管其顺序合并成一组，这种组合称为有重复的组合，其组合种数为 \[ C_n^k = C_{n + k - 1}^k \]
\[ C_n^k = \frac{n}{k}C_{n - 1}^{k - 1} = \frac{{k + 1}}{{n + 1}}C_{n + 1}^{k + 1} = \frac{{k + 1}}{{n - k}}C_n^{k + 1} = \frac{n}{{n - k}}C_{n - 1}^k \] \[ C_n^k = C_n^{n - k} \] \[ C_{n + 1}^k = C_n^k + C_n^{k - 1} \] \[ C_{n + 1}^k = \sum\limits_{j = 0}^k {C_{n - j}^{k - j} } \] \[ C_{n + k + 1}^{n + 1} = \sum\limits_{j = 0}^k {C_{n + j}^n } \] \[ C_{m + n}^k = \sum\limits_{j = 0}^k {C_m^j C_n^{k - j} } \] \[ \begin{array}{l} \sum\limits_{j = 0}^k {C_n^j } = 2^n \\ \left( { - 1} \right)^k C_{n - 1}^k = \sum\limits_{j = 0}^k {\left( { - 1} \right)^j } C_n^j \\ \sum\limits_{j = 0}^n {\left( { - 1} \right)^j C_{2n + 1}^j } = \left( { - 1} \right)^n \frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^2 }} \\ \sum\limits_{j = 0}^n {jC_n^j } = n2^{n - 1} \\ \sum\limits_{j = 0}^n {\left( { - 1\,} \right)^{j + 1} jC_n^j = 0} \\ \sum\limits_{j = 0}^n {j^2 C_n^j = 2^{n - 2} n\left( {n + 1} \right)} \\ \end{array} \] \[ \begin{array}{l} \sum\limits_{j = 1}^k {j^3 C_n^j } = 2^{n - 3} n^2 \left( {n + 3} \right) \\ \sum\limits_{j = 0}^k {\frac{1}{{j + 1}}C_n^j } = \frac{{2^{n + 1} - 1}}{{n + 1}} \\ \sum\limits_{j = 0}^k {\frac{{\left( { - 1} \right)^{j + 1} }}{{j + 1}}C_n^j } = - \frac{1}{{n + 1}} \\ \sum\limits_{j = 0}^k {\frac{{\left( { - 1} \right)^j }}{{j + 1}}C_n^j = \frac{{\left( {2n} \right)!!}}{{\left( {2n + 1} \right)!!}}} \\ \sum\limits_{j = 0}^k {\left( {C_n^j } \right)^2 = \frac{{\left( {2n} \right)!}}{{\left( {n!} \right)^2 }}} \\ \sum\limits_{j = 0}^n {\left( { - 1} \right)^j \left( {C_n^j } \right)^2 } = \left\{ {\begin{array}{*{20}c} {\left( { - 1} \right)^m \frac{{\left( {2m} \right)!}}{{\left( {m!} \right)^2 }},} & {n = 2m} \\ {0,} & {n = 2m + 1} \\ \end{array}} \right. \\ \end{array} \] \[ \begin{array}{l} \sum\limits_{j = 0}^n {j\left( {C_n^j } \right)^2 = \frac{{\left( {2n - 1} \right)!}}{{\left[ {\left( {n - 1} \right)!} \right]^2 }}} \\ \sum\limits_{j = 0}^{n - k} {C_n^j C_n^{k + j} } = C_{2n}^{n - k} = \frac{{\left( {2n} \right)!}}{{\left( {n - k} \right)!\left( {n + k} \right)!}} \\ \sum\limits_{j = 0}^n {C_n^j C_m^j } = C_{n + m}^n \\ \sum\limits_{j = 0}^{\left[ {\frac{n}{2}} \right]} {C_n^{2j} } = 2^{n - 1} \\ \sum\limits_{j = 0}^{\left[ {\frac{{n - 1}}{2}} \right]} {C_n^{2j + 1} } = 2^{n - 1} \\ \sum\limits_{j = 0}^{\left[ {\frac{n}{2}} \right]} {\left( { - 1} \right)^j C_n^{2j} = 2^{\frac{\pi }{2}} } \cos \frac{{n\pi }}{4} \\ \end{array} \] \[ \begin{array}{l} \sum\limits_{j = 0}^{\left[ {\frac{{n - 1}}{2}} \right]} {\left( { - 1} \right)^j C_n^{2j + 1} } = 2^{\frac{n}{2}} \sin \frac{{n\pi }}{4} \\ \sum\limits_{j = 0}^n {\left( {n - 2j} \right)^2 C_n^j = n2^n } \\ \sum\limits_{j = 0}^n {\left( { - 1} \right)^j \left( {n - 2j} \right)^2 C_n^j = \left\{ {\begin{array}{*{20}c} {0,} & {n \ne 2} \\ {8,} & {n = 2} \\ \end{array}} \right.} \\ \sum\limits_{j = 1}^n {C_{\alpha + j - 1}^j = \frac{{\left( {a + 1} \right)\left( {a + 2} \right) \cdots \left( {a + n} \right)}}{{n!}} - 1} \\ \sum\limits_{j = 0}^n {\frac{{\left( { - 1} \right)^j }}{{a + j}}C_n^j } = \begin{array}{*{20}c} {\frac{{n!}}{{a\left( {a + 1} \right) \cdots \left( {a + n} \right)}}} & {} & {\left( {a \ne 0, - 1, - 2, \cdots , - n} \right)} \\ \end{array} \\ \begin{array}{*{20}c} {C_n^{n_1 ,n_2 , \cdots ,n_m } \le m^n } & {} & {\left( {n = n_1 + n_2 + \cdots n_m ,1 \le m \le n} \right)} \\ \end{array} \\ \end{array} \]