The following problem was motivated by the calculation of the dimension of a particular finite dimensional algebra. Fix a positive integer $n$, and consider an $n \times n$ grid of squares. In the outer layer of squares, place a $1$ in each square; in the second-to-outer layer place a $2$ in each square; and so on. For example, when $n=5$ the grid of squares looks like:
1 1 1 1 1
1 2 2 2 1
1 2 3 2 1
1 2 2 2 1
1 1 1 1 1
It is not difficult to show that the sum of the elements in such a square is given by $\binom{n+2}{3}$. However, the presence of a binomial coefficient leads me to wonder if there is a combinatorial reason/proof for this fact, or at least a method which doesn't resort to simply summing the elements.
Let the number in cell counts the # of bricks stacked on that cell. Assume $n$ is odd. Then the lowest layer has $n^2$ bricks, the second lowest layer has $(n-2)^2$ bricks, $\cdots$, and the highest layer has $1$ brick. Therefore, the total number of bricks, or say, the sum of all numbers, is $$ 1^2 + 3^2 + \cdots + (n-2)^2 + n^2 = {n + 2\choose 3} $$ The case for even $n$ is similar.