A fair coin is tossed $$2n$$  times. The probability of getting as many heads in the first $$n$$ tosses as in the last $$n$$ is :

Solution :
The number of possible outcomes of $$2n$$  tosses is $${2^{2n}}.$$  There are $${}^n{C_r}$$ ways of getting $$r$$ heads, with $$0 \leqslant r \leqslant n,$$   in $$n$$ tosses. Therefore, the number of ways of getting $$r$$ heads in both the first $$n$$ and last $$n$$ tosses is $${\left( {{}^n{C_r}} \right)^2}.$$  Summing over all values of $$r$$, the number of favourable ways is
$${\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + ...... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n},$$
So that the required probability is $$\frac{{{}^{2n}{C_n}}}{{{2^{2n}}}}.$$