The problem with this answer is that its not correct to begin with. Assume the problem pertains to two individuals throwing 2 coins then

the total outcomes is indeed $ 1/2^(2n) $ however the amount of possible outcomes is not $\binom{2n}{n}$ so in the case of each having 2 coins the resulting outcomes is 1 head and two tails, this can be done 2 ways $2*2 = 4$

2 heads , this can be done only 1 way $1*1=1$

This is a total of $5$ and $\binom{2*2}{2} = 6$

The problem with this answer is that its not correct to begin with. Assume the problem pertains to two individuals throwing 2 coins then

the total outcomes is indeed $ \frac{1}{2^m} , m= 2n$ however the amount of possible outcomes is not $\binom{2n}{n}$ so in the case of each having 2 coins the resulting outcomes is:

1 head and two tails, this can be done 2 ways $2*2 = 4$

2 heads , this can be done only 1 way $1*1=1$

This is a total of $5$ and $\binom{2*2}{2} = 6$ this is of course not equal and difference of 1 holds as you continue.