Question

In how many ways can the panel be formed if coach $C$ and $F$ should be there together if at all any one of them is there?

Answer 1

Given that panel must consist of both $C\xi F$ or neither of them but not only one of them.

Case $I:$ $C\xi F$ both included.

Only $1$ more position $\xi$ $4$ coaches to choose from $={}^4{C}_1=4.$

Case $II:$ $C\xi F$ both excluded.

$3$ positions $\xi$ $4$ coaches to choose from $={}^4{C}_3=4.$

$\Rightarrow$ Total $=4+4=8$

Hence, the answer is $8.$