Find the number of ways in which 10 players out of 14 players can be selected…

Concept: When certain elements must always be included and certain others must always be excluded from a selection, first remove the excluded elements from the eligible pool and treat the included elements as already placed in the required count. The remaining spots must then be filled by choosing from what remains, using the combination count nCr = n!/(r!·(n − r)).

1. Out of the 14 players, remove the 2 players who must always be excluded, leaving 12 players who are actually eligible for selection.

2. Of the 10 required team spots, 3 are already taken by the players who must always be included, so only 10 − 3 = 7 spots remain open for the rest of the team.

3. Since those 3 included players are already placed, they are also removed from the pool available for the open spots, leaving 12 − 3 = 9 eligible players to fill the remaining 7 spots.

4. The number of ways to fill these 7 open spots from the 9 eligible players is 9C7 = 9!/(7!·2!) = (9 × 8)/(2 × 1) = 36.

Cross-check: Choosing 7 players to include out of 9 is the same as choosing the 2 who are left out, so 9C2 = (9 × 8)/(2 × 1) = 36 as well — the two ways of counting agree.

So the required team can be formed in 36 ways.