(i) Total number of ways of selecting 11 players out of 15 = 15C11
= 15C15-4
= 15C4
\(= \frac{15 \times 14 \times 13 \times 12}{4\times 3 \times 2 \times 1}\)
\(= 1365\)
(ii) When a particular player is always chosen, then we will have to choose 10 players out of remaining 14.
∴ required number of ways = 14C10
= 14C14-10
= 14C4
\(= \frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}\)
\(= 1001\)
(iii) When a particular player is never chosen, then we will have to choose 11 players out of remaining 14.
∴ required number of ways = 14C11
= 14C14-11
= 14C3
\(= \frac{14 \times 13 \times 12}{ 3 \times 2 \times 1}\)
\(= 364\)