[Math] Solving a question about inclusion exclusion principle

combinatoricsinclusion-exclusion

I am trying to solve the following question:

There are $50$ students in a class who are given a test with $3$ questions on it: $Q_1$, $Q_2$, and $Q_3$. All the students answer at least $1$ question. If $12$ students did not answer $Q_1$, $14$ did not answer $Q_2$, $10$ did not answer $Q_3$ and $25$ answered all $3$ questions, then how many students answered exactly $1$ question?

So I define $3$ sets as such:

  • Students who answered $Q_1 = A$
  • Students who answered $Q_2 = B$
  • Students who answered $Q_3 = C$

I am given the following:

  • $|A \cup B \cup C| = 50$
  • $|B \cup C \cap A^c| = 12$ (Students who did not answer $Q_1$)
  • $|A \cup C \cap B^c| = 14$ (Students who did not answer $Q_2$)
  • $|A \cup B \cap C^c)| = 10$ (Students who did not answer $Q_3$)
  • $|A \cap B \cap C| = 25$ (Students who answered all $3$)

The principle of Inclusion-Exclusion states that:
$$|A \cup B \cup C| = |A| + |B| + |C| – |A ∩ B| – |A ∩ C| – |B ∩ C| + |A ∩ B ∩ C|$$

Substituting for the knows values:

$$50 = |A| + |B| + |C| – |A ∩ B| – |A ∩ C| – |B ∩ C| + 25$$

Or $$25 = |A| + |B| + |C| – |A ∩ B| – |A ∩ C| – |B ∩ C|$$

But now I am stuck because the statement above does not give me any information about students who answered exactly 1 question.

How do I move ahead or where did I go wrong?
I dont want to know the answer, I just want to know how to proceed.

Since I got stuck there I went ahead and defined $3$ more sets:

  • Students who answered exactly $1$ question = $D$
  • Students who answered exactly $2$ questions = $E$
  • Students who answered exactly $3$ questions = $F$

I know that $|F| = 25$

and $|D \cup E \cup F| = 50$

But now I am stuck again…

Best Answer

I'd like to flesh out @turkeyhundt's method, which is basically what you've started out writing. Here's a Venn diagram to clarify what variable I'll assign to which question:

enter image description here


We're given the following:

$$a+b+c + x+y+z=25$$

$$b+c+z=12$$

$$a+c+x=14$$

$$a+b+y=10$$

We're trying to find $a+b+c$. If we add the second, third, and fourth equations above we get

$$2a+2b+2c+x+y+z=36$$

Subtracting the first equation above gives us the desired result

$$a+b+c = \boxed{11}$$


Notice that it's not possible to find the individual values of $a$, $b$, and $c$, but it's also not necessary to do that in order to solve the problem.

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