Sets/Venn diagrams question

In a group of 50 students at a summer school, 15 play tennis, 20 play cricket, 20 swim and 7 students do nothing. 3 students play tennis and cricket, 6 students play cricket and swim, 5 students play tennis and swim. How many do all three sports?

Here's what I have:

$n(\mathbb{U}) = 50$
$n(T) = 15$
$n(C) = 20$
$n(S) = 20$

$n(T\cap C) = 3$
$n(C\cap S) = 6$
$n(T\cap S) = 5$
$n(T\cup C\cup S)' = 7$
$n(T\cap C\cap S) = x$

And

$n(\mathbb{U}) = n(T\cap C) + n(C\cap S) + n(T\cap S) + x$
$+ (n(T) - (T\cap C) - (T\cap S) - x)$
$+ (n(C) - (C\cap T) - (C\cap S) - x)$
$+ (n(S) - (S\cap T) - (S\cap C) - x)$
$+ (T\cup C\cup S)'$

Therefore

$50=3+6+5+x$
$+(15-3-5-x)$
$+(20-3-6-x)$
$+(20-6-5-x)$
$+7$

Simplified:
$-2x+41 = 43$
$-2x = 2$
$x=-1$

This answer isn't right because you can't have a negative amount of things. The actual solution to this problem is 2, however this doesn't make sense to me, as:
$3+6+5+2$
$+(15-3-5-2)$
$+(20-3-6-2)$
$+(20-6-5-2)$
$+7$
$=44$
$\ne 50$

• The answer $x=2$ is correct. Have you learned about the principle of inclusion-exclusion? Apr 18, 2017 at 0:26

This is just begging for a venn diagram, so here goes:

So we have that: $n(\mathbb{U}) = 50$
$n(T) = 15$
$n(C) = 20$
$n(S) = 20$

Then from the venn diagram, it should be easy to see:

$50 = 20+15+20+7 -(3-x+5-x+6-x)-2x$

$\implies x =2$ as required

• I understand everything until -2x. Isn't AUBUC = n(A)+n(B)+n(C) - n(A∩B)-n(A∩B)-n(A∩C) + n(A∩B∩C) ?
– Ivan
Apr 18, 2017 at 0:56
• @usernamesAreHard Well just think about it in terms of the venn diagram - when you add $n(C)+n(T)+n(S)$, you are counting $x=n(C\cap T\cap S)\quad3$ times, so we must subtract it twice. Do you follow? Apr 18, 2017 at 0:57
• Ahh, I understand. Thanks for the help.
– Ivan
Apr 18, 2017 at 1:00
• @usernamesAreHard No problem. By the way the formula you posted is correct, but we're not using it in that 'form' as we're using the venn diagram instead. Apr 18, 2017 at 1:02