Three meetings where each attends exactly two

Question

Suppose three meetings of a group of professors were arranged in Mumbai, Delhi and Chennai. Each professor of the group attended exactly two meetings. $21$ professors attended Mumbai meeting, $27$ professors attended Delhi meeting, $30$ professors attended Chennai meeting. How many of them attended both Chennai and Delhi meetings?

Now, letting $M,D,C$ be set of professors who attended Mumbai, Chennai and Delhi meetings respectively, by inclusion-exclusion principle, we have

$$n(M\cup D\cup C)=n(M)+n(D)+n(C)-n(M\cap D)-n(M\cap C)-n(D\cap C)+n(M\cap D\cap C)$$

where $n(M\cap D\cap C)=0$, $n(M),n(D),n(C)$ are known, and not $n(M\cup D\cup C), n(M\cap D)$ and $n(M\cap C)$. Hence I concluded that it is not possible find $n(D\cap C)$.

But the given condition that each professor attended exactly two meetings is making me doubt if my conclusion is right. Can someone help me?

Thanks in advance.

Answer 1

Let the number who attended each pair of conferences be $CD, CM, DM$ then you are given $$CD+CM=30$$$$CD+DM=27$$$$CM+DM=21$$

You are told that every professor attended two conferences, so no professor attended only one and none attended all three. Three equations. Three unknowns. You should be able to solve.

Answer 2

Venn diagram to help understand Mark Bennet's answer easier

Answer 3

$CD+CM=30$

$CD+DM=27$

$CM+DM=21$

Solving we have: $CD-DM=9$

$CD+DM=27$

So we have

$2 CD=36$

$CD=18$