Inclusion-Exclusion Principle Problem for 3 Sets Where Intersections are Unknown 
In a class of 30 children, 20 take Latin, 14 take Greek, and 10 take
  Hebrew. If no child takes all three languages and eight children take
  no language, how many children take Greek and Hebrew?

So we have 3 sets:
$$L, G, H$$
We are given that no child takes all three languages:
$$|L \cap G \cap H| = 0$$
And that eight children take no language:
$$|U| - |L \cup G \cup H| = 8$$
Where $U$ is the universal set (class of 30 children; $|U| = 30$). By simple algebra we can find $|L \cup G \cup H| = 22$. So here are my known quantities:
$$|L| = 20$$
$$|G| = 14$$
$$|H| = 10$$
$$|L \cap G \cap H| = 0$$
$$|L \cup G \cup H| = 22$$
By the inclusion-exclusion principle we can set up the following:
$$|L \cup G \cup H| = |L| + |G| + |H| - |L \cap G| - |L \cap H| - |G \cap H| + |L \cap G \cap H|$$
Plugging in for known quantities:
$$22 = 20 + 14 + 10 - |L \cap G| - |L \cap H| - |G \cap H| + 0$$
$$|L \cap G| + |L \cap H| + |G \cap H| = 22$$
Okay. But how can I find $|G \cap H|$ from this?
 A: Let's just think about it. We've got $30$ kids, but $8$ of them are taking no languages, so forget about them; we've got $22$ kids taking languages.
Of those $22$ kids, $14$ take Greek and $10$ take Hebrew, $14+10=24,$ so we've got at least $2$ kids taking Greek and Hebrew.
Since nobody takes three languages, the kids taking Greek and Hebrew can't be taking Latin. Since $20$ of the $22$ kids are taking Latin, that means at most $2$ are taking Greek and Hebrew.
Conclusion: exactly $2$ kids taking Greek and Hebrew.
A: So here's what Kenneth and I came up with by playing with the numbers using a Venn diagram. There are only a few ways to distribute the 22 students who take 2 classes amongst the three intersections. 
First note $|L\cap G|+|L\cap H|+|G\cap H|=22$ and $|L\cap G \cap H|=0$ as above.
Also $|L\cap H|+|G\cap H|\le |H|=10$ because both intersections are in $H$. Putting this into $|L\cap G|+|L\cap H|+|G\cap H|=22$, we get $|L\cap G|\ge 12$.
$|L\cap G|+|G\cap H|\le |G|=14$ implies $|L\cap G|\le 14$.
Combining, we have $12\le |L\cap G|\le14$. Also note $|L\cap G|+|L\cap H|\le 20$.
If $|L\cap G|=14$, then $|G\cap H|=0$ and $|L\cap H|=10$, but that's impossible as it would mean $|L\cap G|+|L\cap H|=24$. If $|L\cap G|=13$, then $|G\cap H|=0\  or\ 1$ and $|L\cap H|=9\ or \ 8$ respectively. This is impossible as it would mean $|L\cap G|+|L\cap H|=22\ or \ 21$ respectively. 
Finally we reach $|L\cap G|=12$. In this case, $|G\cap H|=0\,\ ,1\   or\ 2$ and $|L\cap H|=10,\ 9,\ or \ 8$ respectively. The first two cases again make $|L\cap G|+|L\cap H|$ too large, leaving the final possibility, $|L\cap G|=12$, $|G\cap H|=2$, and $|L\cap H|=8$ which works and is the only case which does.
All this finally gives the answer: two students take both Greek and Hebrew.
