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:

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

  • $\begingroup$ 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 $F\subset E$. $\endgroup$ – amWhy 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? $\endgroup$ – bp99 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 $|F| =[ E\cap F| = 10$ $\endgroup$ – amWhy Jan 18 '17 at 18:01
  • 1
    $\begingroup$ It is enough to conclude that $U = E \cup (U\setminus E)$, having found that $|U| = |E| \cup |U\setminus (E\cup F)| = |E|\cup |U\setminus E|$. $\endgroup$ – amWhy 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 $10$? $\endgroup$ – bp99 Jan 18 '17 at 18:11

@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 $10$.

  • $\begingroup$ 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... $\endgroup$ – bp99 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 $\endgroup$ – bp99 Jan 21 '17 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.