Combinatorics: problem with sets We have set $A=\left \{a_1,a_2,...a_{11}  \right \} $. How many ordered pairs $(X,Y)$, $X\subset A$, $Y\subset A$, have property $|X|=8$, $|Y|=7$, $|X\cap Y|=5$?
Thanks in advance!
 A: Well how many ways can you pick 5 out of 11? There's $\frac{11!}{5!6!}$, then we need to pick 5 more numbers from the remaining 6, there's 6 ways to do this, and 3 will be in $X$ and 2 will be in $Y$, there's $\frac{5!}{3!2!}$ ways to do this. So the final solution is $$\frac{11!}{5!6!} \cdot 6 \cdot \frac{5!}{3!2!}$$
A: We can look at the problem this way.
First we will construct the set $X$. This can be done in C$(11,8)$ different ways. Now we will construct the set $Y$. We have to be careful here because there are restrictions on the elements that $Y$ can have. Namely, $Y$ and $X$ must have 5 common elements.
That's why we will first put 5 elements that are already in $X$, into $Y$. This can be done in C$(8,5)$ ways. Now, we must fill up $Y$ with two elements that are not in $X$. The number of these elements is $11-8=3$. Thus, we can construct $Y$ in C$(8,5)\cdot$C$(3,2)$ ways.
Using the multiplication principle, we get that the number of ordered pairs is C$(11,8)\cdot$C$(8,5)\cdot$C$(3,2)$.
