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I've stumbled upon the following challenging question. Find a closed formula for the following summation $$ S(a,b,n) = \sum_{k=0}^n {a \choose k} {b \choose n-k}$$ for all possible parameters $a,b$.

Anyone happens to see how to solve this one?

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marked as duplicate by amWhy, Davide Giraudo, azimut, Old John, Najib Idrissi Nov 16 '13 at 16:17

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5 Answers 5

up vote 13 down vote accepted

Since ${a\choose k}$ is the coefficient of $x^k$ in the polynomial $(1+x)^a$ and ${b\choose n-k}$ is the coefficient of $x^{n-k}$ in the polynomial $(1+x)^b$, the sum $S(a,b,n)$ of their products collects all the contributions to the coefficient of $x^n$ in the polynomial $(1+x)^a(1+x)^b=(1+x)^{a+b}$.

This proves that $S(a,b,n)={a+b\choose n}$.

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A counting argument:

If there are $a$ items of type A and $b$ items of type B, then

$$\sum_{k=0}^{n} {a \choose k} {b \choose n-k}$$

is the number of ways to choosing $n$ items from them: choose $k$ of type A and $n-k$ of type B, vary $k$ from $0$ to $n$ and add up.

Thus the sum you seek is $${a+b \choose n}$$

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1  
This is the best answer so far. Anyone who doesn't understand it will learn something valuable by coming to understand it. –  Michael Hardy Jun 27 '11 at 0:56
    
+1,this is indeed the best answer! –  VelvetThunder Sep 5 '11 at 19:45

One way to solve this is by using generating series. With generating series, we barely have to think about what we are doing, and arrive at the answer nicely. Consider $$\sum_{n=0}^{\infty}\sum_{k=0}^{n}\binom{a}{k}\binom{b}{n-k}x^{n}.$$

Switching the order of summation, this becomes

$$\sum_{k=0}^{\infty}\binom{a}{k}\sum_{n=k}^{\infty}\binom{b}{n-k}x^{n}=\sum_{k=0}^{\infty}\binom{a}{k}x^{k}\sum_{n=0}^{\infty}\binom{b}{n}x^{n}.$$ Then, by the binomial theorem this is $$\left(1+x\right)^{a}\left(1+x\right)^{b}=(1+x)^{a+b}.$$

Since the $n^{th}$ coefficient of this is $\binom{a+b}{n}$, we see that $$\sum_{k=0}^n\binom{a}{k}\binom{b}{n-k}=\binom{a+b}{n}.$$

Hope that helps,

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The book A=B
http://www.math.upenn.edu/~wilf/AeqB.html

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