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The exercise goes like this:

In a class, there are 32 people. They learn different languages. 25 people learn English, 10 people learn French. We also know that 7 people learn neither of the two. How many people learn both languages?

This is not a hard exercise, I would simply draw a Venn-diagram like this: venn-diagram sketch

And then I can easily find an equation to solve, knowing that there are 32 people in total: $$ (25-x)+(x)+(10-x)+(7)=32 \\ 42-x=32 \\ x=10 $$

That's fine; As I said, this is an easy exercise. My question is, how would I be able to solve such a problem differently? For example, by only using set theory identities, or something, starting with:

$$ |\mathbb{U}| = 32 \; \land |E| = 25 \; \land |F| = 10 \; \land | \overline{E \cup F} | = 7 \\ | E \cap F|=? $$

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  • $\begingroup$ Do you see that $$ \vert U\vert = \vert E\vert + \vert \overline{E\cup F}\vert.$$ Also note that if $x = 10,$ as you initially found, we have that $F\subset E$. $\endgroup$ – Namaste Jan 18 '17 at 17:42
  • $\begingroup$ @amWhy Yes, at least I can see that the equation is true, although I also understand how the second deduction is true. But that didn't answer my question; I am wondering whether I can solve the problem without drawing diagrams, or using any creativity, as if this was an algebraic exercise? $\endgroup$ – bertalanp99 Jan 18 '17 at 17:48
  • $\begingroup$ Your diagram does not depict the situation at hand. I used nothing more that the realization that, since $32 = 27 +7$, the cardinality of F is irrelevant, and therefore, it must be a subset of E so that $E\cap F \subset E$ and such that $|F| =[ E\cap F| = 10$ $\endgroup$ – Namaste Jan 18 '17 at 18:01
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    $\begingroup$ It is enough to conclude that $U = E \cup (U\setminus E)$, having found that $|U| = |E| \cup |U\setminus (E\cup F)| = |E|\cup |U\setminus E|$. $\endgroup$ – Namaste Jan 18 '17 at 18:11
  • $\begingroup$ You are right, but one could not know that before realising that $F$ is a subset of $E$, n'est-ce pas? I understand that after noticing that $ |U| = |E| + |U \backslash E | $, we can deduce that $ |F| = |E \cap F| $. But how do we know that that equals $10$? $\endgroup$ – bertalanp99 Jan 18 '17 at 18:11
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@amWhy gave a solution in the comments, but what I was looking for exactly is this 'process':

$$ |\mathbb{U}|=32 \:;\: |E|=25 \:;\: |F|=10 \:;\: |\overline{A \cup F}|=7 \\ |\mathbb{U}|=|E \backslash F|+|E \cap F|+|F \backslash E|+|\overline{E \cup F}|=15+7+|E \cap F| \\ |E \cap F|=32-15-7=10 \\ |E \cap F|=10 $$

So the answer is $10$.

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  • $\begingroup$ This process is basically identical to your original method, but it's much harder to verify and to see what's going on. $\endgroup$ – Patrick Stevens Jan 21 '17 at 12:46
  • $\begingroup$ I don't understand, why would it be harder to verify? I know that it's the same method, but it's somehow different, I can't explain properly why... $\endgroup$ – bertalanp99 Jan 21 '17 at 13:20
  • $\begingroup$ I have an intuition for space and number. I have very little intuition for formulae and numbers. $\endgroup$ – Patrick Stevens Jan 21 '17 at 14:30
  • $\begingroup$ @PatrickStevens I am sorry, I don't understand you or your point $\endgroup$ – bertalanp99 Jan 21 '17 at 14:45

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