Let S be a Set and let A be a subset of S. how many options there are to choose 2 subsets from S that their intersection is exactly A?

Question

I'm struggling with this combinatoric problem - i marked the size of Set S as n and the size of A as k. I first thought of it this way: in order that the interaction of 2 subsets of S will be exactly A . the two subsets need to hold all A elements both. after that i left with n-k elements i can divide\or not to the two subsets i created . but i am not allowed to put the same element from the n-k elements in both sets ( otherwise the interaction wont be A). so i decide to choose i subsets from n-k that 0<=i<=n-k and then also to choose j subsets from n-k-i elements that left. i am counting twice so i divided my answer by 2 . but i left with an answer that include "sum" and i need a final answer without the sum operator. so this is what i achieved so far:
$$\sum_{i=0}^{n-k}({n-k \choose i}\sum_{j=0}^{n-k}{n-k \choose j}$$ this answer divide by 2 and i am getting my answer. but please help me find a way to find a final answer without sum operator.or maybe a different perspective over this question

Answer 1

As you say, both subsets have to include $A$ so start there. Then each of the $n-k$ other elements of $S$ can go in one subset, the other, or neither, for three choices. That would give $3^{n-k}$ ordered ways to choose the subsets. They are ordered because we started by saying one was the first subset.

If you care about unordered pairs of subsets you have to note that the one choice of not adding any elements to either subset is symmetric but all the others can be interchanged to produce another ordered pair, so we divide the nonsymmetric pairs by $2$ and add back in the symmetric pair, giving $$\frac 12\left(3^{n-k}-1\right)+1=\frac 12\left(3^{n-k}+1\right)$$