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?

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.
