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The question may have multiple correct answers

A. ${}^{16}{{C}_{11}}$

B. ${}^{16}{{C}_{7}}$

C. ${}^{16}{{C}_{9}}$

D. ${}^{20}{{C}_{9}}$

Answer

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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.