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Let $X_n$ be the set of all word of the length $2 n$ over the alphabet $\{A,B\}$ which contain as many A's as B's.

The amount of elements of $X_n$ is $\displaystyle \binom{2n}{n}$, but why?

I thought about it for a long time but really am a bit slow today. Does anybody have a (simple) combinatorical explanation why this applies?

Thanks in advance!

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The number of words of length $n$ with $k$ $A$s is ${n \choose k}$. Do you know how to prove this? What interpretations of the binomial coefficients are you familiar with? – Qiaochu Yuan Jun 25 '11 at 20:49
@qiaochu-yuan As I know the binomial coefficient is the amount of possibilities of building k-element subsets of an original set containing n elements. Your definition can be easily proved when considering the subsets as the indices where an A is inserted. – muffel Jun 25 '11 at 21:06
Cool. So, consider a word of length $n$ with $k$ $A$s. How can you convert this into a specification of a $k$-element subset of a set of size $n$? – Qiaochu Yuan Jun 25 '11 at 21:10
@qiaochu-yuan a word can be representated by its indices (1,2,3,$\cdots$,n), and a k -element subset is just that set of indices that may turn into an A. Thanks! – muffel Jun 25 '11 at 21:14
up vote 6 down vote accepted

There are $2n$ positions to be filled in a word of length $2n$. Once you know which $n$ positions are filled with $A$'s, the other $n$ positions must contain $B$'s, so the word is completely determined. There are ${2n} \choose {n}$ ways to choose which $n$ positions get the $A$'s, so there are ${2n} \choose {n}$ such words.

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@brian-m-scott just great, thanks! – muffel Jun 25 '11 at 21:07

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