2 particular adults cannot be together in a selection

Question

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

addition of 2 will give my answer.

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 ?

Answer 1

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$.

Answer 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 $2$.