Okey this one is easy(if you seen this one once) We have to show: $${n \choose 0}^2+{n \choose 1}^2+...+{n \choose n}^2={2n \choose n} $$$${n \choose 0}^2+{n \choose 1}^2+\cdots+{n \choose n}^2={2n \choose n} $$ Define a set A, $$A={{a_1,...,a_n,a_{n+1},...a_{2n}}}$$$$A=\{a_1,\ldots,a_n,a_{n+1},\ldots,a_{2n}\}$$ consisting of 2n$2n$ - Elements. Now we declare V$V$, is a set of n$n$-sets of A$A$. Obviously the cardinality of V, $$|V|={2n \choose n}$$.$V,$ $$|V|={2n \choose n}.$$ Since one can choose n$n$ Elements from 2n$2n$ Elements in $${2n \choose n}$$ ways.
Okey now we have the right side. The for the left side you have give a disjunctive partition of the V$V$ set. This should be done in the following way: $V_i$has has exactly i$i$-Elements from ${a_1,..,a_n}$${a_1,\ldots,a_n}$, then the cardinality $|V_i|={n \choose i}{n \choose n-i}={n \choose i}{n \choose i}={n \choose i}^2$. Then with the Rule of Sum we have the left side. And we are done. The Tricky Part ist to create the Partition to use the rule of sum.
