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How to prove this?

$n$, $m$ and $k \in \mathbb{N_{0}} $

$$\binom{n+m}{k} = \sum_{i=0}^{k} \binom{m}{i} \binom{n}{k-i}$$

Martin Sleziak
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Mohbenay
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  • You can prove it yourself using the identity $\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$ and of course by checking out the various links here and on Wikipedia. – StubbornAtom Jan 22 '17 at 13:12

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Say you have $n+m$ numbered balls, where $n$ are red, and $m$ are blue. Then $\binom{n+m}{k}$ is the number of ways to choose $k$ balls regardless of colour, while $\binom{m}{i} \binom{n}{k-i}$ is the number of ways to choose $i$ blue balls and $k-i$ red ones from the same collection. Sum over all possible $i$, and you have the number of ways to pick $k$ balls regardless of colour. Thus the two expressions count the same thing, and must therefore be equal.

Arthur
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  • Yeah thank you , actually your answer was the answer of the question before this one , and now the question wants to solve it using other resolving way – Mohbenay Jan 22 '17 at 12:02
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    What's wrong with a combinatorial proof? And if there are solutions you don't want, you should say that in the question itself, preferably before posting. That way, you save people like me a few minutes of typing up an answer you don't want. – Arthur Jan 22 '17 at 12:18