As here they are talking about choosing n elementsout of 2n elements. so they can choose 1 element out of 2n or 2 elements out of 2n.....................2n elements out of 2n.

so total no of cases 2nc1 + 2nc2 + 2nc3+.............+2nc2n= 2^(2n)= 4^n

ways of getting n heads out of 2n tries = ^{2n}C_{n}

probability of getting exactly n heads and n tails = (1/2^{n})(1/2^{n})

number of ways = ^{2n}C_{n}/4^{n}