Question

Let $X=\{1,2,3,...,10\}$. Find the the number of pairs $(A,B)$ such that $A$ $\subseteq$ $X$, $B$ $\subseteq$ X, $A\ne B$ and $A\cap B=\{5,7,8\}$.

Answer 1

Let $A$ $\cup$ $B=Y$, $B$\$A=M$, $A$\$B=N$ and $X$\$Y=L$.
Then $X$ is the disjoint union of $M,N,L$ and $A$ $\cap B$.
We have, $A$ $\cap$ $B=\{5,7,8\}$ is fixed.
The remaining seven elements $1,2,3,4,6,9,10$ 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 $A=B=\{5,7,8\}$ and this case has to be omitted.
Therefore total number of pairs $(A,B)$ such that $A\subseteq X$, $B\subseteq X$, $A\ne B$ and $A\cap B=\{5,7,8\}$ is $3^7-1$.