An inclusion exclusion type problem In a group of 265 people, 200 like singing, 110 like dancing and 55 like painting. 60 like both singing and dancing, 30 like both singing and painting and 10 like all three activities. How many persons like only dancing and painting. 
So let $S$ be the set of singers, $D$ the set of dancers and $P$ the set of painters. Therefore $$|S|+|D|+|P|-|S\cap D|-|S \cap P|-|D \cap P|+|S \cap D \cap P|=265$$ Now substituting the values
$$200+110+55-60-30- |D \cap P|+10=265$$ gives $|D \cap P|=20$.
This seems correct to me, but the answer given is $|D \cap P|=10$. Is there something wrong with my reasoning?
 A: The formula you cite gives $|S\cup D\cup P|$, not the total number of people, so your claim that it should equal $265$ is unfounded. In fact the numbers given do not allow computing the requested $|\overline S\cap D\cap P|$; setting $n=|\overline S\cap\overline D\cap\overline P|$ gives $|\overline S\cap D\cap P|=10+n$, where $n$ can be any number with $0\leq n\leq 15$.
$$\begin{array}{|c|c|c|c|}\hline
S&D&P&\text{number}\\\hline
0&0&0&n\\\hline
0&0&1&15-n\\\hline
0&1&0&40-n\\\hline
0&1&1&\color{red}{10+n}\\\hline
1&0&0&120\\\hline
1&0&1&20\\\hline
1&1&0&50\\\hline
1&1&1&10\\\hline
\end{array} $$
A: You have found $|D\cap P|$ correctly, but you're asked to find $|D\cap P \cap \text{not }S|$.
A: We are going to use the implicit assumption that: The number of people who don't like either dancing or painting or singing is zero. Without this assumption the problem does not have a unique solution.
I just hope that it is easy to interpret the following figures.
We can immediately infere the following:

Based on the number of singers we can proceed:

Now, given that the number of non-dancers is $155$ we can proceed:

The number of non-painters is $210$ so:

From the number of non-singers ($65$) follows:

Now, we can tell that the number of those who like only dancing and painting is $10$. 
