HMMT November, 2018

Solution:
There are $6 \cdot 6$ ordered pairs of integers between $1$ and $6$ inclusive and $21$ unordered pairs of integers ($\binom{6}{2} = 15$ different pairs and $6$ doubles). Then, $A = \binom{36}{2} + 36 = 666$ and $B = 21 \cdot 21 = 441$. Therefore $A - B = 225$.

For general $n$, there are $n^{2}$ ordered pairs of integers and $\frac{n(n+1)}{2}$ unordered pairs of integers. Then $A = \frac{n^{2}(n^{2}+1)}{2}$ and $B = \frac{n^{2}(n+1)^{2}}{4}$ so
$$A - B = \frac{n^{2}\left(2\left(n^{2}+1\right)-(n+1)^{2}\right)}{4} = \left(\frac{n(n-1)}{2}\right)^{2}$$