# A “fast” approach to compute $\sum_{i=0}^{n} \binom{19}{i} \times \binom{7}{n-i}$ [duplicate]

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How to find a closed formula for the given summation

I am looking for a fast/best approach to compute $$\sum_{i=0}^{n} \binom{19}{i} \times \binom{7}{n-i}$$

For example,if $n=4$ the answer is $\binom{26}{4}$.Please explain your appraoch.

ADDED: After some experimentation with different numbers I think in general,$\sum_{i=0}^{n} \binom{p}{i} \times \binom{q}{n-i} = \binom{p+q}{n}$ holds.Does this result correct? If yes how could we proof this result.Any ideas?

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## marked as duplicate by Zev ChonolesSep 5 '11 at 20:03

Guess the rule then prove it. – Yuval Filmus Sep 5 '11 at 19:33
– Zev Chonoles Sep 5 '11 at 19:35
@ Zev Chonoles:Thanks for the link :-) – VelvetThunder Sep 5 '11 at 19:40
@Zev Chonoles:You may delete this question,I think the other question has already provided me with sufficient information that one could possible want to know about this problem.Thanks. – VelvetThunder Sep 5 '11 at 20:02
@FoolForMath: I don't think there's a need to delete it, I will just close as duplicate. – Zev Chonoles Sep 5 '11 at 20:03

This is just a way of finding out how many ways there are to choose n things from 26 objects. So for any n, the answer will be $26 \choose n$. How? Say we divide the 26 things into 2 groups, call 19 of them 'red' and 7 'blue.' Then we sum over the possibilities of taking 0 blue, n red or 1 blue, n-1 red, ... , n blue, 0 red. That is exactly your sum, and that is why it's just $26 \choose n$.