Each of the 30 students in a class plays at least one of squash, hockey and tennis.
18 students play squash
19 students play hockey
17 students play tennis
8 students play squash and hockey
9 students play hockey and tennis
11 students play squash and tennis
Find the number of students who play all three sports.
I've tried working backwards, putting $x$ in the middle, then $8-x$, $9-x$ and $11-x$ further outwards, but I can't seem to deduce the outermost circles. I wonder what I'm missing?
While your approach would work for two sets, it does not work for three sets.
Consider the following diagram.
Suppose we wish to count $|A \cup B \cup C|$. If we simply add the number of elements in each set, we will count each element that is in two sets twice, once for each set to which it belongs. Since we only want to count such elements once, we must subtract the number of elements that are in two sets from the total. However, if we do so, we count each element that is all three sets three times when we add the number of elements that are in each set, once for each set to which it belongs, then subtract it three times when we subtract the number of elements that are in two sets, once for each of the $\binom{3}{2} = 3$ pairs of sets to which it belongs. Therefore, we have not counted such elements at all. We want to count them once, so we add them to the total. Doing so yields $$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|$$ This is the Inclusion-Exclusion Principle for three sets.
Let $S$ be the set of students who play squash, $H$ be the set of students who play hockey, and $T$ be the set of students who play tennis. We are given \begin{align*} |S \cup H \cup T| & = 30\\ |S| & = 18\\ |H| & = 19\\ |T| & = 17\\ |S \cap H| & = 8\\ |S \cap T| & = 11\\ |H \cap T| & = 9 \end{align*} You wish to find $|S \cap H \cap T|$, which you can determine using the equation above by setting $A = S$, $B = H$, and $C = T$.