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I need to prove the following: ${n\choose m}={n\choose n-m}$

With the definition: ${n\choose m}= \left\{ \begin{array}{ll} \frac{n!}{m!(n-m)!} & \textrm{für \(m\leq n\)} \\ 0 & \textrm{für \(m>n\)} \end{array} \right.$

and $n,m\in\mathbb{N}$.

I'm stuck at how to even start this. Using induction? Any help would be appreciated, I tried to search SA, but couldn't find an answer.

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  • $\begingroup$ Substitute $m = n-m'$ in your definition then. $\endgroup$ – peterwhy Nov 27 '14 at 16:15
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Hint: Just use factorial formula.

$${n\choose m}= \frac{n!}{m! \cdot (n-m)!}$$


Solution:

$${n\choose n-m} = \frac{n!}{(n-m)! \cdot (n -(n-m))!} = \frac{n!}{(n-m)! \cdot m!} = \frac{n!}{m! \cdot (n-m)!} = {n\choose m}$$

I believe, rest is obvious.

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An other way:

to choose $m$ balls in a box that contain $n$ balls is equivalent that to not choose the $n-m$ other balls.

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$$ {n\choose n-m} = \frac{n!}{(n-m)!(n-(n-m))!} = \frac{n!}{(n-m)!m!} = {n \choose m} $$

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