Total propositional functions (distinct) such that their truth tables have equal number of true and false?

Question

I had found a below question

How many $n$-variable propositional functions (distinct) are there such that their truth tables have equal number of true and false?

My attempt: Since there are $2^n$ possible truth values therefore total such functions are $C(2^{n-1},0)^2+C(2^{n-1},1)^2 + C(2^{n-1},2)^2+...+C(2^{n-1},2^{n-1})^2$ But answer given is $C(2^{n},2^{n-1})$. Where I am wrong? And if I am correct then is there any better way to approach this solution?

Answer 1

From Vandermonde's identity

$\sum_{k=0}^{2^{n-1}}$ $2^{n-1}\choose{k}$$=$$2^{n}\choose{2^{n-1}}$

Let me know if you have some more direct way to approach this problem where right hand expression could be discovered directly without mathematical manipulations.

Answer 2

Both answers are correct, since

$\binom{2^{n-1}}{0}^{2}+\binom{2^{n-1}}{1}^2 + \binom{2^{n-1}}{2}^2+\dots+\binom{2^{n-1}}{2^{n-1}}^2 = \binom{2^{n}}{2^{n-1}}$

For example, when $n=3$:

$1^2 + 4^2 + 6^2 + 4^2 + 1^2 = 1+16+26+16+1= 70 = \binom{8}{4}$