今天学习组合数

定理1:上指标求和

\[\sum^n_{i=0}\dbinom{m+i}{i}=\dbinom{m+n+1}{m+1} \]

证明

\[\begin{aligned} \sum^n_{i=0}\dbinom{m+i}{i}&=\sum^n_{i=0}\dbinom{m+i}{m}\\ &=\dbinom{m}{m}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &=\dbinom{m+1}{m+1}+\dbinom{m+1}{m}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &=\dbinom{m+2}{m+1}+\dbinom{m+2}{m}+\dots+\dbinom{m+n}{m}\\ &\dots\\ &=\dbinom{m+n+1}{m+1} \end{aligned} \]

定理2

\[\sum_{i=0}^{n} i\dbinom{n}{i}=n2^{n-1} \]

证明

\[\begin{aligned} \sum_{i=0}^{n} i\dbinom{n}{i} &= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{n-1}+n\dbinom{n}{n}\\ &= 0\dbinom{n}{0}+1\dbinom{n}{1}+\dots+(n-1)\dbinom{n}{1}+n\dbinom{n}{0}\\ &= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{1}+\frac{n}{2}\dbinom{n}{0}\\ &= \frac{n}{2}\dbinom{n}{0} +\frac{n}{2}\dbinom{n}{1}+\dots+\frac{n}{2}\dbinom{n}{n-1}+\frac{n}{2}\dbinom{n}{n}\\ &= \sum_{i=0}^n \frac{n}{2}\dbinom{n}{i} = \frac{n}{2} \sum_{i=0}^n \dbinom{n}{i}\\ &= \frac{n}{2}2^n = n2^{n-1} \end{aligned} \]

定理3

\[\dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}=n:m:n-m \]

证明

\[\begin{aligned} \dbinom{n}{m}&=\dfrac{n^\underline{m}}{m^\underline{m}}\\ \dbinom{n-1}{m-1}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}\\ \dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\ \dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dfrac{(n-1)^\underline{m-1}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{m^\underline{m}}\\ &=\dfrac{m^\underline{m}}{(m-1)^\underline{m-1}}:\dfrac{(n-1)^\underline{m}}{(n-1)^\underline{m-1}}\\ &=m:n-m\\ \dbinom{n}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}&=\dbinom{n-1}{m-1}+\dbinom{n-1}{m}:\dbinom{n-1}{m-1}:\dbinom{n-1}{m}\\ &=n:m:n-m \end{aligned} \]

推论

\[\dfrac{\dbinom{n-1}{m-1}}{m}=\dfrac{\dbinom{n}{m}}{n} \]