Question

The number of ways in which a team of eleven players can be selected from 22 players, always including 2 of them and excluding 4 of them is
The question may have multiple correct answers
A. ${}^{16}{{C}_{11}}$
B. ${}^{16}{{C}_{7}}$
C. ${}^{16}{{C}_{9}}$
D. ${}^{20}{{C}_{9}}$

Answer 1

Hint: To solve this question, we should know the details about the non-fixed places in the team and the players who are always excluded or included. We can infer that 2 out of the 22 players should be selected always into the team and 4 players out of the 22 players are always not selected in the team. In total, there are 9 (positions excluding the two positions for the two players) for $22-2-4=16$ players. The number of ways of selecting r players out of n players is given by ${}^{n}{{C}_{r}}$. Using this formula, we can get the answer.

Complete step-by-step solution:
In the question, it is given that a team of 11 players is to be selected out of 22 players. We can observe that there are constraints that 2 of the 22 players should be selected into the team and 4 of the 22 players should not be selected.
To tackle this problem, we should divide the selection into 2 parts. The first part is selecting the two compulsory players into 2 positions of the team and they are subtracted from the 22 players. The four players are excluded from the selection and they are subtracted from 22 players too. The second set of selection is done for $22-2-4=16$ players for the remaining $11-2=9$ positions.
We can write the combinations for the first part as
$N={}^{2}{{C}_{2}}=\dfrac{2!}{2!\left( 2-2 \right)!}=1$ where ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
The number of ways of selecting r players out of n players is given by ${}^{n}{{C}_{r}}$
For the second part, there are 16 players for 9 positions, we can write the number of possible ways as n=16, r=9
$M={}^{16}{{C}_{9}}$
The total number of ways is given by multiplying the values of M and N as we have done both the events.
Total ways = $M\times N={}^{16}{{C}_{9}}\times 1={}^{16}{{C}_{9}}$
We also know the relation that ${}^{n}{{C}_{r}}={}^{n}{{C}_{n-r}}$
Here n=16 and r=9, we get
${}^{16}{{C}_{9}}={}^{16}{{C}_{16-9}}={}^{16}{{C}_{7}}$
$\therefore $The number of ways of selecting a 11 player team is ${}^{16}{{C}_{9}}\text{ or }{}^{16}{{C}_{7}}$. The answers are options B and C.

Note: Students can make a mistake by not considering the constraints given in the question. That leads to a selection of 11 players out of 20 players which can be written as ${}^{20}{{C}_{11}}={}^{20}{{C}_{20-11}}={}^{20}{{C}_{9}}$ which leads to a wrong option-D.