范德蒙恒等式

范德蒙恒等式(英文：Vandermonde's Identity)是一个有关组合数的求和公式。 ( n + m k ) = ∑ i = 0 k ( n i ) ( m k − i ) {\displaystyle {\binom {n+m}{k}}=\sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}} 甲班有 m {\displaystyle m} 个同学，乙班有 n {\displaystyle n} 个同学，从两个班中选出 k {\displaystyle k} 個同學有 ( n + m k ) {\displaystyle {\binom {n+m}{k}}} 种方法。 从甲班选 k − i {\displaystyle k-i} 名，从乙班选 i {\displaystyle i} 名有 ( n i ) ( m k − i ) {\displaystyle {\binom {n}{i}}{\binom {m}{k-i}}} 种方法，考虑所有情况 i = 0 , 1 , … , k {\displaystyle i=0,1,\ldots ,k} ，从两个班中合計 k {\displaystyle k} 选出個同學有 ∑ i = 0 k ( n i ) ( m k − i ) {\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}} 种方法。 所以 ( n + m k ) = ∑ i = 0 k ( n i ) ( m k − i ) {\displaystyle {\binom {n+m}{k}}=\sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}} 注意到 ( 1 + x ) n ( 1 + x ) m = ( 1 + x ) n + m {\displaystyle (1+x)^{n}(1+x)^{m}=(1+x)^{n+m}} 等號左邊化簡成 ( 1 + x ) n ( 1 + x ) m = ( ∑ i = 0 n ( n i ) x i ) ( ∑ j = 0 m ( m j ) x j ) = ∑ k = 0 m + n ( ∑ i = 0 k ( n i ) ( m k − i ) ) x k {\displaystyle (1+x)^{n}(1+x)^{m}=\left(\sum _{i=0}^{n}{\binom {n}{i}}x^{i}\right)\left(\sum _{j=0}^{m}{\binom {m}{j}}x^{j}\right)=\sum _{k=0}^{m+n}\left(\sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}\right)x^{k}} 等號右邊則根據定義 ( 1 + x ) n + m = ∑ k = 0 n + m ( n + m k ) x k {\displaystyle (1+x)^{n+m}=\sum _{k=0}^{n+m}{\binom {n+m}{k}}x^{k}} 比較 x k {\displaystyle x^{k}} 係數，可得 ( n + m k ) = ∑ i = 0 k ( n i ) ( m k − i ) {\displaystyle {\binom {n+m}{k}}=\sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}} ∑ k i j ( n 1 k 11 , k 12 , … , k 1 t ) … ( n s k s 1 , k s 2 , … , k s t ) = ( n 1 + n 2 + ⋯ + n s r 1 , r 2 , … , r t ) {\displaystyle \sum _{k_{ij}}{n_{1} \choose k_{11},k_{12},\dots ,k_{1t}}\dots {n_{s} \choose k_{s1},k_{s2},\dots ,k_{st}}={n_{1}+n_{2}+\dots +n_{s} \choose r_{1},r_{2},\dots ,r_{t}}} 其中 ( n n 1 , n 2 , … , n m ) = n ! n 1 ! n 2 ! … n m ! , k 1 l + k 2 l + ⋯ + k s l = r l , l = 1 , … , t {\displaystyle {n \choose n_{1},n_{2},\dots ,n_{m}}={\frac {n!}{n_{1}!n_{2}!\dots n_{m}!}},k_{1l}+k_{2l}+\dots +k_{sl}=r_{l},l=1,\dots ,t} 展开 ( x 1 + x 2 + ⋯ + x t ) n 1 + n 2 + ⋯ + n s = ( x 1 + x 2 + ⋯ + x t ) n 1 … ( x 1 + x 2 + ⋯ + x t ) n s {\displaystyle (x_{1}+x_{2}+\dots +x_{t})^{n_{1}+n_{2}+\dots +n_{s}}=(x_{1}+x_{2}+\dots +x_{t})^{n_{1}}\dots (x_{1}+x_{2}+\dots +x_{t})^{n_{s}}} 可得以上结论。 范德蒙恒等式是超几何函数的一个整数特例。 2 F 1 ( a , b ; c ; 1 ) = ∑ n = 0 ∞ a ( n ) b ( n ) c ( n ) n ! = Γ ( c ) Γ ( c − a − b ) Γ ( c − a ) Γ ( c − b ) , ℜ ( c ) > ℜ ( a + b ) {\displaystyle {}_{2}F_{1}(a,b;c;1)=\sum _{n=0}^{\infty }{\frac {a^{(n)}b^{(n)}}{c^{(n)}n!}}={\frac {\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}},\quad \Re (c)>\Re (a+b)} ∑ i = 0 k ( n i ) ( m k − i ) = m ! k ! ( m − k ) ! ∑ i = 0 ∞ ( − n ) ( i ) ( − k ) ( i ) ( m − k + 1 ) ( i ) i ! = m ! k ! ( m − k ) ! 2 F 1 ( − n , − k ; m − k + 1 ; 1 ) {\displaystyle \sum _{i=0}^{k}{\binom {n}{i}}{\binom {m}{k-i}}={\frac {m!}{k!(m-k)!}}\sum _{i=0}^{\infty }{\frac {(-n)^{(i)}(-k)^{(i)}}{(m-k+1)^{(i)}i!}}={\frac {m!}{k!(m-k)!}}{}_{2}F_{1}(-n,-k;m-k+1;1)} = m ! k ! ( m − k ) ! Γ ( m − k + 1 ) Γ ( n + m + 1 ) Γ ( n + m − k + 1 ) Γ ( m + 1 ) = ( n + m ) ! k ! ( n + m − k ) ! = ( n + m k ) {\displaystyle ={\frac {m!}{k!(m-k)!}}{\frac {\Gamma (m-k+1)\Gamma (n+m+1)}{\Gamma (n+m-k+1)\Gamma (m+1)}}={\frac {(n+m)!}{k!(n+m-k)!}}={\binom {n+m}{k}}} 李松槐 杨伏香. 用数学模型证明范得蒙(Vandermonde)恒等式. 河南教育学院学报(自然科学版). 1999, (2) [2015-09-20]. （原始内容存档于2020-01-15）.  Hac`ene Belbachir. A combinatorial contribution to the multinomial Chu-Vandermonde convolution (PDF). RECITS Laboratory. 2014 [2018-06-12]. （原始内容 (PDF)存档于2020-11-30）.  Bailey, W.N. Generalized Hypergeometric Series (PDF). Cambridge University Press. 1935 [2018-06-12]. （原始内容 (PDF)存档于2017-06-24）.