In this $4*4$ matrix, we only have choices for the elements enclosed in the red rectangle.
The corresponding blue rectangle would have same elements as of the red rectangle.
We got choices for $4+3+2+1$ elements.
We can extend this to $n*n$ matrix, in which we'd have choices for $n+(n-1)+(n-2)+...+1$ elements.
=> $1+2+3+...+n$ elements
=> $\frac{n(n+1)}{2}$ elements
Two choices for $\frac{n(n+1)}{2}$ elements means $2^{\frac{n(n+1)}{2}}$
Option C