Question

In how many ways can a football team of 11 players be selected from 16 players ?

How many of these will (i) include 2 particular players ? (ii) exclude 2 particular players ?

Answer 1

No of players = 16

No. of players to be selected = 11

$\therefore$ No of combination ${=}^{16}{C}_{11}$

$=\frac{16!}{11!5!}=\frac{16\times 15\times 14\times 13\times 12}{5\times 4\times 3\times 2}=4368$

(i) Include 2 particular players

$\Rightarrow$ Now we have to select 9 more out of remaining 14

$\therefore$ Required number of ways ${=}^{14}{C}_9$

$=\frac{14!}{9!5!}=\frac{14\times 13\times 12\times 11\times 10}{5\times 4\times 3\times 2}$

$=2002$

(ii) Exclude 2 particular players $\to$ now we have to select 11 players out of 14 players

${=}^{14}{C}_{11}=\frac{14!}{11!3!}=\frac{14\times 13\times 12}{3\times 2}$

$=364$