Let set B be of cardinality n.
Total subsets(A) possible are : nC0 + nC1 + nC2 + ... + nCn. i.e nCr number of subsets exist with r cardinality.
Note that for each r, summation has $2^{n-r}$ terms to sum.
Case 1: r= 0. Which is $\Phi$ .
Total terms = $2^{n}$.
Total terms when |C\A| even = nC0 + nC2 + nC4 + ... + nC(n-1) { if n is odd, nCn otherwise }
Similarly for odd = 2ⁿ - |C\A|
Even contributes to 1 whereas odd contributes to -1.
Therefore Summation = 0 as 2ⁿ terms are present with half as 1 & half as -1.
Case 2: r = 1 , total terms = $2^{n-1}$
Total terms when |C\A| even = nCr + nC(r+2) + ... + nCn { if n is odd n-1 otherwise }
Total terms when |C\A| odd = nC(r+1) + nC(r+3) + ...
Again we are end up with total even terms with half contributing to 1 & half -1.
same situation will arise for every r ≠ n.(as for such r, $2^{r}$ is always even) i.e summation = 0 for all r, r ≠ n.
Case n: r = n , total terms = $2^{n-n}$ = 1. This is the case when B = A.
THEREFORE |C\A| = 0 as both are equal.
Summation = 1.
therefore answer is : 1 if A = B, 0 otherwise.
Lemme know if I'm wrong.