How many ways can a football team of 11 players be selected from 16 players how many of them will I include two particular players ii exclude two particular players?

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.

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

 exclude 2 particular players?

If 2 particular players are excluded, it would mean that out of 14 players, 11 players are selected.  Required number of ways =\[{}^{14} C_{11} = \frac{14!}{11! 3!} = \frac{14 \times 13 \times 12}{3 \times 2 \times 1} = 364\]

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