Possible Duplicate:
Proof that $\binom{n}{\smash{0}}^2+\binom{n}{1}^2+\cdots+\binom{n}{n}^2=\binom{\smash{2}n}{n}$ using a counting argument
It is an exercise in a book on discrete mathematics.It is like this.
Give a combinatorial argument to prove that $\sum_{k=0}^{n}C\left ( n,k \right)^{2}=C\left ( 2n,n \right )$.
I don't know how to begin with it.It's hard for me to construct such a set,or the context.I learned it by myself and no one around can help me whit my answer. Thanks for your help.
Consider the right hand side of the equality. What does it tell you? Translated to english the RHS is "the number of n-subsets of a 2n-set" (where by n-set I mean a set of cardinality n). Now consider the $2n$-set. Split it up in two sets of equal size $n$. You can pick a n-set by choosing $n$ elements from one of the sets and zero from the other, or choosing $n-1$ elements from one of the n-sets and $1$ element from the other and so on. I.e. $$C(2n,n)=\sum{C(n,k)C(n,n-k)=\sum{C(n,k)^2}}$$ The last equality follows since the number of ways of choosing $k$ elements from a $n$ set is the same number of ways as choosing those $n-k$ elements that shall not belong to a $k$-subset.
Use the Chu-Vandermonde identity, as recommended by t.b.: $$ {m+n \choose r}=\sum_{k=0}^r{m \choose k}{n \choose r-k},\qquad m,n,r\in\mathbb{N}_0, $$ and set $m=n$, $r=n$ and use ${n \choose n-k}={n \choose k}$ to get: $$ \sum_{k=0}^n{n \choose k}^2= {2n \choose n}$$
