Combinatorial sum II

This is a very standard proof exercise through binomial expansion. Consider $(1+x)^{2n}=(1+x)^n(1+x)^n$ $\sum_{k=0}^{2n}\dbinom{2n}{k}x^k=[\dbinom{n}{0}+\dbinom{n}{1}x+...+\dbinom{n}{n}x^n][\dbinom{n}{0}+\dbinom{n}{1}x+...+\dbinom{n}{n}x^n]$

Consider the coefficient of $x^n$ on both sides, we have $\dbinom{2n}{n}=\dbinom{n}{0}\dbinom{n}{n}+\dbinom{n}{1}\dbinom{n}{n-1}+...+\dbinom{n}{n}\dbinom{n}{0}$

Since $\dbinom{n}{r}=\dbinom{n}{n-r} \,\forall\, 0\le r \le n$ , we have $\dbinom{n}{0}^2+\dbinom{n}{1}^2+...+\dbinom{n}{n}^2=\dbinom{2n}{n}$

Therefore $a=2n$ , $b=n$ , and $\frac{a}{b}=\frac{2n}{n}=\boxed{2}$

Notice that $\left( \begin{array}{c} n\\ k\\ \end{array} \right) = \left( \begin{array}{c} n\\ n-k\\ \end{array} \right)$ . So $\left( \begin{array}{c} n\\ k\\ \end{array} \right)^2 = \left( \begin{array}{c} n\\ k\\ \end{array} \right) \cdot \left( \begin{array}{c} n\\ n-k\\ \end{array} \right)$ . The sum of all those terms gives the number of way or extract $n$ balls of a set of $2n$ balls separated in two subsets of $n$ balls each.