HMMT February

We consider the sets drawn in a Venn diagram. Note that each element that is in at least one of the subsets lies in these seven possible spaces. We split by casework, with the cases based on $N=\left|R_{7}\right|=|A \cap B \cap C|$. Case 1: $N=2$ Because we are given that $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=0$. But we also know that $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, so $\left|R_{1}\right|=2$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=2$. Since these regions are distinguishable, we multiply through and obtain $\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}=2520$ ways. Case 2: $N=1$ In this case, we can immediately deduce $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=1$. From this, it follows that $\left|R_{1}\right|=4-1-1-1=1$, and similarly, $\left|R_{2}\right|=\left|R_{3}\right|=1$. All seven regions each contain one integer, so there are a total of $(8)(7) \ldots(2)=40320$ ways. Case 3: $N=0$ Because $\left|R_{4}\right|+N=\left|R_{5}\right|+N=\left|R_{6}\right|+N=2$, we must have $\left|R_{4}\right|=\left|R_{5}\right|=\left|R_{6}\right|=2$. Since $\left|R_{1}\right|+\left|R_{5}\right|+\left|R_{6}\right|+N=4$, we immediately see that $\left|R_{1}\right|=0$. Similarly, $\left|R_{2}\right|=\left|R_{3}\right|=0$. The number of ways to fill $R_{4}, R_{5}, R_{6}$ is $\binom{8}{2}\binom{6}{2}\binom{4}{2}=2520$. This clearly exhausts all the possibilities, so adding gives us $40320+2520+2520=45360$ ways.