A team of 4 students is to be selected from 12 students with conditions conditions-
two particular students refuse to be together.
other two students wish to be together only.
Approach-total number of ways to form team without any restrictions is $$\binom{12}{4}$$. Now I subtract those combinations in which the students who refuse to be together are together i.e is $$\binom{10}{2}$$
Now I subtract the combinations in which the  willing students are not together i.e one of them is there $$2\binom{11}{3}$$ So doing subtractions i get $120$ as answer but the answer is $226$. What I am doing wrong?

This problem relates to the Principle of Inclusion and Exclusion. You subtract the cases where $A$ and $B$ are together (although they don't want to) and $C$ and $D$ are not together (although they insist on participation together) twice. So you have to add them again. That results in $2*\binom{8}{1}$ possibilities. Also, you have to subtract $2\binom{10}3$, because you can't choose the person that is in the group ($C$), nor the person he wants to be with ($D$). 
