Question

In how many ways can a football team of 11 players be selected from 16 players? How many of these will include 2 particular players?
A. 4638,2020
B. 4658,2001
C. 4368,2002
D. None of these

Answer 1

This problem deals with permutations and combinations. But here a simple concept is used. Although this problem deals with combinations only. Here factorial of any number is the product of that number and all the numbers less than that number till 1.
$\Rightarrow n! = n(n - 1)(n - 2).......1$
The no. of combinations of $n$ objects taken $r$ at a time is determined by the formula which is used:
$\Rightarrow {}^n{c_r} = \dfrac{{n!}}{{\left( {n - r} \right)!r!}}$

Complete step-by-step answer:
Given that there is a football team.
We have to select 11 players from the given 16 players.
The no. of ways in which 11 players can be selected from 16 players, is given below:
$\Rightarrow {}^{16}{c_{11}} = \dfrac{{16!}}{{\left( {16 - 11} \right)!11!}}$
$\Rightarrow {}^{16}{c_{11}} = 4368$
Now we have to find that from obtained no. of ways that, how many of these will include 2 particular players.
Before we had to select 11 players, but here as already 2 players are selected, hence the remaining players to be selected are 9 players.
So we are selecting 9 players from 14 players.
The no. of ways in which 9 players can be selected from 14 players is given by:
$\Rightarrow {}^{14}{c_9} = \dfrac{{14!}}{{\left( {14 - 9} \right)!9!}}$
$\Rightarrow {}^{14}{c_9} = 2002$

Final Answer: The no. of ways a football team of 11 players be selected from 16 players and the no. of ways in which these players will include 2 particular players are 4368 and 2002 respectively.

Note:
Please note that while selecting the no. of players in which 2 particular players should be included, here to 2 particular players for sure, we need to keep those 2 players aside, as they have to be selected. Now the remaining no. of players from 16 players after 2 players got selected are 14 players. That means we have to choose from the remaining 14 players.