0
$\begingroup$

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 !!

|cite|improve this question
$\endgroup$
  • $\begingroup$ @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 ? $\endgroup$ – Math Newbie May 12 '17 at 20:28
  • $\begingroup$ Ah I see what your problem is. See my answer below. $\endgroup$ – kccu May 12 '17 at 20:43
  • $\begingroup$ 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. $\endgroup$ – N.Bach May 12 '17 at 20:50
  • $\begingroup$ @N.Bach it was on an old exam and that is the exercise.. Poor guys that were solving it haha $\endgroup$ – Math Newbie May 13 '17 at 4:38
2
$\begingroup$

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: enter image description here

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 $|A^c \cap B \cap C|=-2$.)

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Thaaaanks alot !! Seems to be bugging me but now it's perfectly understandable 😆 $\endgroup$ – Math Newbie May 12 '17 at 20:45

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.