Question

FairPlay Athletics club has six coaches, viz, $A,B,C,D,E$ and $F$. A panel of coaches comprising of three members has to be formed In how many ways can the panel be formed if coach $E$ and $B$ cannot be together on the panel?

Answer 1

Given that $E\xi B$ cannot be together in the $3$ member panel.

$\Rightarrow E\xi B$ not together $=$ Total $-$ $E\xi B$ together

$\Rightarrow$ Total combinations $={}^6{C}_3=5\times 4=20.$

Combinations with $E\xi B$ together.

$\Rightarrow$ Only $1$ remaining slot to be filled by $4$ coaches $={}^4{C}_1=4$

$\Rightarrow$ Required $=20-4=16$

Hence, the answer is $16.$