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I need to prove that :

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

without expanding.

I started by seperating the left hand side to get:

$$\binom{m+M}{n} = \binom{m}{n}\binom{M}{n}$$, and am trying to open up the right hand side with Pascal's identity but I am not sure how to make it work out to the required equality.

Vivid
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    Whit a combinatorial argument: chosing $n$ people among $m+M$ people is equivalent to chose $i$ people in the first $m$ and then $n-i$ in the $M$ left. – MoebiusCorzer Nov 02 '15 at 16:05
  • "I started by separating the left hand side..." No, $\binom{n+M}{n}$ is not equal ot $\binom{m}{n}\binom{M}{n}$. – Thomas Andrews Nov 02 '15 at 16:08

2 Answers2

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Given a set of $m+M$ items, separated into sets of $m$ and $M$. The ways of choosing a subset of $n$ items is the number of ways of choosing $0$ items from the first set and all $n$ from the second, plus the number of ways of choosing $1$ item from the first and $n-1$ from the second...

Chris Kerridge
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To get $n$ items from 2 big piles of $m$ and $M$, you can either compute this directly getting $\binom{m+M}{n}$ or you can split by how many items you get from the $m$ pile, but then you must sum over all such possibilities, getting $$\sum_{k=0}^n \binom{m}{i} \binom{M}{n-i}$$

gt6989b
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