Number of ways to select $10$ from $20$ people so that $5$ can play guitar and $5$ can play violin In a class of $20$ people, all of them can either play violin or guitar. Given that $16$ if them can play guitar and $11$ of them can play violin, how many ways are there to select $10$ people from that group of people such that $5$ of them can play guitar and $5$ of them can play violin?
I have drawn a venn diagram for this question and I found that people who play only guitar is $9$, play both guitar and violin is $7$ and play only violin is $4$. So for the combination, should I just $16\choose5$ for the people who play guitar and $11\choose5$ for the people who play violin? Or instead of that, I should used $6\choose5$ for those who play violin? since the guitar part already taken $5$? 
 A: I am not able to see a slick way by which it can be done at one stroke.
Let us label only guitar/violin players as $G/V$ and the versatile ones as $X$. A workable way is:
Consider various numbers of X in the violin group, and form the guitar group from the residue.
$5X: \binom75 * \binom{11}5$
$4X: \binom74\binom41 *\binom{12}5$
..........
$1X: \binom71\binom44 * \binom{15}5$
Finally, add up.
A: Let $x_1$ be the number of strict guitarists selected, $x_2$ be the number of guitarist/violinists in the guitar group, $x_3$ be the number of guitarist/violinists in the violin group, and $x_4$ be the number of strict violinists selected.  Then we want to count the number of solutions in non-negative integers to the following set of equations:
$$(1)\qquad \qquad x_1+x_2 = 5\\
(2)\qquad \qquad x_3+x_4=5
$$
subject to $x_1 \le 9$, $x_4 \le 4$, and $x_2+x_3 \le 7$.
Our strategy is to drop the constraint $x_2+x_3 \le 7$ and then divide the solutions to (1) and (2), with the constraints $x_1 \le 9$ and $x_4 \le 4$, into two groups: those with $x_2+x_3 \le 7$ and those with $x_2+x_3 \ge 8$.
If we consider (1) and (2) with $x_1 \le 9$ and $x_4 \le 4$, it is clear there are $6$ choices for $x_1$ and $5$ choices for $x_4$, for $6 \times 5 = 30$ solutions in all.
Now consider (1) and (2) with the added constraint $x_2+x_3 \ge 8$.  We introduce a new non-negative integer variable $y$ with $x_2+x_3 = 8 + y$.  Subtracting (1) and (2) from this last equation, we have
$$(3) \qquad \qquad x_1+x_4+y=2$$
Any solution to (3) corresponds to a unique solution to (1) and (2), since given $x_1$ and $x_4$ we can find $x_2$ and $x_3$.  It is clear the constraints $x_1 \le 9$ and $x_4 \le 4$ are non-binding in (3) and can be ignored, so the number of solutions to (3), through a standard stars-and-bars computation, is $\binom{4}{2}=6$.
Suppose $N$ is the number of solutions to the original problem, including the constraint $x_2+x_3 \le 7$.  Then combining the results above,
$$N + 6 = 30$$
Solve for $N$.
