There are $6 \cdot 6$ ordered pairs of integers between 1 and 6 inclusive and 21 unordered pairs of integers \left(\binom{6}{2}=15\right.$different pairs and 6 doubles). Then,$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 $A=\frac{n^{2}\left(n^{2}+1\right)}{2}$ and $B=\frac{n^{2}(n+1)^{2}}{4}$ so $$A-B=\frac{n^{2}\left(2\left(n^{2}+1\right)-(n+1)^{2}\right)}{4}=\left(\frac{n(n-1)}{2}\right)^{2}$$