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I need to solve this binomial summation but cant seem to get it using binomial identities I learnt in school and college first-year:

$$S=\sum_{i=q}^{p-q}{\binom{i}{q}}{\binom{n-i}{p-q}}$$ p,q,n are positive integers such that $q\leq p\leq n$

Am I missing out something naive and clever?

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It is probably better to call this a sum, and not a series. – 1015 Feb 2 '13 at 13:52
So $p-q\geq q$ is another assumption here? – 1015 Feb 2 '13 at 13:57
no the only assumptions are listed above. – Shagun Feb 2 '13 at 13:58
sorry for confusing you with this thing actually the summation is to be done for all listed values of i and q need not be less than p-q – Shagun Feb 2 '13 at 13:59
Do you mean that the sum is actually $\sum_{i=p-q}^q$ if $p-q<q$? – 1015 Feb 2 '13 at 14:01

I guess it is $nCp$. Explanation : In the process of selecting p students from a group of n students ,You divide the group of n students into two groups. One having i students and other having n-i students. Now you select q students from the i students and rest p-q sudents from the rest students.As i would run from q to p-q(considering q<=p-q) and you sum them you will get the no. of ways in which you can select p students from n students.

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For $n=4,p=2,q=1$ you get $\binom11\binom31=1\times3=3\neq6=\binom42$ – Marc van Leeuwen Apr 22 '13 at 13:59

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