2019/20 AHSMC Part 1 Problem 9

The number of dominos having different number of dots on each square is $100 + 99 + \cdots + 2 + 1 = \dfrac{100 \cdot 101}{2} = 5050$ while the number of dominos that have the same number on each half is $101$, hence the total number of dominos is $5151$. The answer is (C).

Alternate solution: If a square can contain anywhere from $0$ to $n$ dots, then there are $$\binom{n+1}{2} = \frac{(n+1)!}{(n-1)! \, 2!} = \frac{(n+1)n}{2}$$ dominos with different numbers of dots on each half, plus $n+1$ dominos that have the same number on each half. Hence, there are $$\frac{(n+1)n}{2} + n + 1 = \frac{n^2 + 3n + 2}{2}$$ dominos in total. When $n = 100$, we have $$\frac{100^2 + 3 \cdot 100 + 2}{2} = 5151$$ dominos.