Given that we need to choose 11 players for a team out of available 16 players,

Let us assume the choosing the no. of ways be N,

⇒ N = choosing 11 players out of 16 players

⇒ N = ¹⁶C₁₁

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N = 4368 ways

(i) It is told that two players are always included.

It is similar to selecting 9 players out of the remaining 14 players as 2 players are already selected.

Let us assume the choosing the no. of ways be N₁,

⇒ N₁ = choosing 9 players out of 14 players

⇒ N₁ = ¹⁴C₉

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₁ = 2002 ways

(ii) It is told that two players are always excluded.

It is similar to selecting 11 players out of the remaining 14 players as 2 players are already removed.

Let us assume the choosing the no. of ways be N₂,

⇒ N₂ = choosing 11 players out of 14 players

⇒ N₂ = ¹⁴C₁₁

We know that

And also n! = (n)(n – 1)(n – 2)…………2.1

⇒

⇒

⇒

⇒ N₂ = 364 ways

∴ The required no. of ways are 4368, 2002, 364.