Question

find the number of ways in which 6 player out of 11 players can be selected so as to include 3 particular players.


i want full solution bzzz this is frm permutation and combination

Answer 1

You have to choose 6 players out of 11 players

But 3 particular players must be included. So you only have to choose 3 players from rest 8 player.

3 players can be chosen from 8 players in 8C3 ways.

$\text{Number of ways} = {}^{8} C_3\\ \\ = \frac{8!}{3! \times (8-3)!} \\ \\ = \frac{8!}{3! \times 5!} \\ \\ =\frac{8 \times 7 \times 6 \times 5!}{3 \times 2 \times 1 \times 5! } \\ \\ =56 \\ \\ \boxed{ \red{ \textbf{Number of ways = 56}}}$

Answer 2

Total number of players=11
Number of players to be selected = 6

Now 3 particular player must be in the team.

total number of ways =8C3
$= \frac{8factorial}{3factorial \times 5factorial} \\ = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \\ = 8 \times 7 \\ = 56$