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Please provide a combinatorial proof for the following:

Prove the identity $$\sum_{i=0}^{k}{m+k-i-1 \choose k-i}{n+i-1 \choose i}={m+n+k-1 \choose k}$$

Hint: use idea of "selection with repetition".

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Do you mean $$\sum_{i=0}^k\binom{m+k-i-1}{k-i}\binom{n+i-1}{i}=\binom{m+n+k-1}{k}\quad?$$ What did you try to do? – Martin Jan 29 '13 at 12:23
Yes I that's the question. For the R.S I said that its counts the no. of ways to select k objects from m+n objects with repetition, but am not sure how to interpret the summation side. – Sarah Jan 29 '13 at 12:37
It looks like a homework problem. – Joe Z. Jan 30 '13 at 3:51
@JoeZeng: or an exercise in a book. It's hard to second guess. In cases like this, without further information, we usually accept what the author says. – robjohn Feb 4 '13 at 2:09
up vote 5 down vote accepted

Imagine you have $m$ red objects and $n$ blue objects, and you want to select $k$ objects with possible repetitions. This can be done (as you stated yourself in the comment) in


different ways.

On the other hand, among these $k$ selected objects, there could be exactly $k$ red ones and $0$ blue ones, there could be $k-1$ red and $1$ blue, etc. Each of these cases are mutually exclusive, and there are

$$\binom{m+(k-i)-1}{k-i} \binom{n+i-1}{i}$$

ways to choose exactly $k-i$ red objects and $i$ blue objects. Sum over all $i$ and you get your identity.

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thank you for your help! – Sarah Jan 29 '13 at 13:39
@Sarah: You're welcome! And since you're new here, don't forget to upvote answers and questions you find useful (and not just answers to your own questions). If you're really happy with a particular answer, you can "accept" it, but it's usually a good idea to wait awhile before doing that -- even better answers can appear later. – mrf Jan 29 '13 at 13:43
Yes for sure! But how do I upvote? – Sarah Jan 29 '13 at 13:47
@Sarah: Look at the grey arrows in the left column. Clicking the up-arrow gives an upvote, and likewise, the down-arrow gives a downvote. The number attached to each post is the difference between upvotes and downvotes. – mrf Jan 29 '13 at 13:49

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