Q:

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 !! 

A:

Actually there are no sets which satisfy the given conditions.
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 $|A^c \cap B \cap C|=-2$.)

