$\begin{array}{1 1}(A)\;16C_{11}\\(B)\;14C_9\\(C)\;16C_9\\(D)\;16C_2\end{array} $

- $C(n,r)=\large\frac{n!}{r!(n-r)!}$

Total no of players =16

Hence we have to select 11 players out of 16

So required no of ways =$11C_{16}$

Since 2 particular players are always included,so we have to select 9 players out of remaining 14 players

This can be done in $14C_9$ ways

$\Rightarrow 14C_9=\large\frac{14!}{9!\times 6!}$

$\Rightarrow \large\frac{14\times 13\times 12\times 11\times 10\times 9!}{91\times 6\times 5\times 4\times 3\times 2\times 1}$

$\Rightarrow 333.66$ ways

Hence (B) is the correct answer.

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