# Confused - Inclusion exclusion principle

Hello to everyone who sees this question.. I have seen a problem and it was looking very easy so I tried to solve it but something seems to go wrong..

If there are 3 sets A, B, and C where |A ∪ B ∪ C| = 34, |C ∩ B ∩ A| = 2,
|A ∩ B| = 4, |A ∩ C| = 4, |A| = 16, |B| = 7, |C| = 17. What is the value of |B ∩ C| ?


I have tried by the Inclusion-Exclusion principle

|A ∪ B ∪ C| = |A| + |B| + |C| - |A ∩ B| - |A ∩ C| - |B ∩ C| + |A ∩ B ∩ C|
->   34 = 16 + 7 + 17 - 4 - 4 - |B ∩ C| + 2
->   34 = 34 - |B ∩ C|
->    0 = |B ∩ C|


However it is stated that |C ∩ B ∩ A| = 2.. Thanks !!

• @kccu I dont know where I'm wrong it comes up that |B \cap C| = 0 and |A \cap B \cap C| = 2.. How can it be ? – Math Newbie May 12 '17 at 20:28
• Ah I see what your problem is. See my answer below. – kccu May 12 '17 at 20:43
• Where did you find this exercise ? The only explanations I can think of are (1) you made an error while copying the exercise or (2) the problem statement is erroneous. – N.Bach May 12 '17 at 20:50
• @N.Bach it was on an old exam and that is the exercise.. Poor guys that were solving it haha – Math Newbie May 13 '17 at 4:38

Note that $A \cap B = (A \cap B \cap C) \cup (A \cap B \cap C^c)$, and these are disjoint. Since $|A\cap B|=4$ and $|A\cap B \cap C|=2$, we must have $|A\cap B \cap C^c|=2$.
Similarly we find $|A\cap B^c\cap C| = 2$.
Let $x$ denote $|A^c \cap B \cap C|$. Then we get the following venn diagram:
We are supposed to have $|A \cup B \cup C|=34$, i.e., \begin{align*} 10+2+2+2+x+(3-x)+(13-x)&=34\\ 32-x &= 34\\ x&=-2. \end{align*} This is impossible, so there are no such sets. (Note this is consistent with finding that $|B\cap C|=0$ and $|A \cap B \cap C|=2$... these would imply $|A^c \cap B \cap C|=-2$.)