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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:
$3+6+5+2$
$+(15-3-5-2)$
$+(20-3-6-2)$
$+(20-6-5-2)$
$+7$
$=44$
$\ne 50$

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

1 Answer 1

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

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

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  • $\begingroup$ 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) ? $\endgroup$
    – Ivan
    Apr 18, 2017 at 0:56
  • $\begingroup$ @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? $\endgroup$
    – mrnovice
    Apr 18, 2017 at 0:57
  • $\begingroup$ Ahh, I understand. Thanks for the help. $\endgroup$
    – Ivan
    Apr 18, 2017 at 1:00
  • $\begingroup$ @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. $\endgroup$
    – mrnovice
    Apr 18, 2017 at 1:02

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