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?
 A: Your first step is good--subtract $_{10}C_2$. However, the next thing you subtract should be $2*_{10}C_3$, not $2*_{11}C_3$. You don't actually have $11$ choices for the other three members of the team, as you want only one of the two "willing students" to be on the committee; the other is "eliminated" from being selected, leaving only $10$ options. Finally, there is some overlap between the two groups that you're subtracting. You must add back in the instances that you've double-subtracted. These are the teams that include both non-willing students and only one of the willing students. There are $2*_8C_1$ of these!
As others have mentioned, this process of adding back in what gets subtracted twice is an illustration of the inclusion/exclusion principle.
A: 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 ($D$). 
A: This is a cool example of inclusion/exclusion principle.
You start with every thing.
Subtract those you do not want
So far you are doing right. But the problem is you may subtract the same answer twice. So you have to put them back. My guess is you need to add back those that you subtracted twice.
Sorry, hit submit too soon!
Let $X_T$ be all the choices. $X_1$ choices that violate the first condition, $X_2$ the choices that violate the second. Let $X_B$ be the choices that violate both. Then the number of choices is
$$ \left|X_T\right| - \left|X_1\right|- \left|X_2\right|+ \left|X_B\right|$$
So find $\left|X_B\right|$ and add it back.
A: This isn't a direct answer to your question, but it gives another approach that could be used:
If we call the two people who refuse to be together A and B, and the two people who will only be on the team together C and D, then we can divide this into 2 cases:
1) If we select C and D and not both of A and B, there are $\binom{10}{2}$ ways to choose the other two people, and only one of these involves selecting A and B; so there are $\binom{10}{2}-1=44$ possibilities in this case.
2) If we select neither C nor D and not both of A and B, there are $\binom{10}{4}$ ways to select 4 people 
without C and D, and $\binom{8}{2}$ of these selections involve choosing both A and B; so there are 
$\binom{10}{4}-\binom{8}{2}=210-28=182$ possibilities in this case.
Therefore the total number of teams is $44+182=226$.
