# 2 particular adults cannot be together in a selection

A 4 member committee is to be formed from a group of 9 adults.Find the number of ways this committee can be formed if 2 particular adults cannot be together.

my attempt:

Case 1- N(both not in committee) = 7C4 case 2- N(one in and one out) = 7C3

However, the mistake in my step is in case 2-

N(one in and one out) = 7C3 X 2

why must i multiply by 2 ?

Because there could be two possibilities in case $$2$$ , for example if person $$A$$ has been selected in a team then remaining ways would be $$7 \choose 3$$ . However person $$B$$ can also be selected in the same way , therefore you need to multiply the answer by $$2$$.
Therefore final answer is $$7\choose 4$$ $$+$$ $$7\choose 3×2$$.
Person $$1$$ and person $$2$$ cannot be together. In your case $$2$$, you calculated, for instance, person $$1$$ in and person $$2$$ out. However, there's the other way around, namely person $$2$$ in and person $$1$$ out. That's why you should multiply by $$2$$.