# Prove that the combination formula can be reduced to…

Prove that:

$$\frac{m!}{k!(m-k)!} = \frac{m}{k}\frac{m-1}{k-1}\cdots\frac{m-k+1}{1}$$

It's quite obvious when I write down some terms, but I just don't know how to make a rigorous proof. Any hints or help are appreciated!

Thanks.

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$$\frac{m!}{k!(m-k)!} =\frac{1\cdot\ 2\cdot...\cdot(m-k)\cdot(m-k+1)\cdot...\cdot m}{(1\cdot\ 2\cdot...\cdot k)(1\cdot\ 2\cdot...\cdot(m-k))}$$
\begin{align*} \frac{n!}{k!(n-k)!}&=\frac{\prod_{i=1}^ni}{\left(\prod_{i=1}^ki\right)\left(\prod_{i=1}^{n-k}i\right)}\\\\ &=\frac{\left(\prod_{i=1}^{n-k}i\right)\left(\prod_{i=n-k+1}^ni\right)}{\left(\prod_{i=1}^ki\right)\left(\prod_{i=1}^{n-k}i\right)}\\\\ &=\frac{\prod_{i=n-k+1}^ni}{\prod_{i=1}^ki}\\\\ &=\frac{\prod_{i=1}^k(n-k+i)}{\prod_{i=1}^ki}\\\\ &=\prod_{i=1}^k\frac{n-k+i}i \end{align*}
$$\frac{m!}{k!(m-k)!}=\binom{m}{k}=\frac{m}{k}\binom{m-1}{k-1}=\frac{m}{k}\frac{m-1}{k-1}\binom{m-2}{k-2}=\frac{m}{k}\frac{m-1}{k-1}\frac{m-2}{k-2}...\frac{m-(k-1)}{1}$$