Question

Let $X$ = {1, 2, 3,.,.,., 12}. Find the the number of pairs {A, B} such that $A\subseteq X,B\subseteq X,A\ne B$ and $A\cap B$ = {2, 3, 5, 7,8}

Answer 1

Let $A\cup B=Y$, B\A = M, A\B = N and X\Y = L. Then X is the disjointunion of M, N, L and $A\cap B$.
We have, $A\cap B$ = {2, 3, 5, 7, 8} is fixed. The remaining seven elements 1, 4, 6, 9, 10, 11, 12 can be distributed in any of the remaining sets M, N, L.
This can be done in $3^7$ ways.
Of these if all the elements are in the set L, then A = B = {2, 3, 5, 7, 8} and this case has to be omitted.
Hence the total number of pairs {A, B} such that $A\subseteq X,B\subseteq X,A\ne B$ and $A\cap B=2,3,5,7,8$ is $3^7-1$