$\begin{array}{1 1}(A)\;3860\\(B)\;3960\\(C)\;3660\\(D)\;3760\end{array} $

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Four bowlers can be selected from the five players and seven players can be selected from 12 players (17-5)=12 is the number of ways of selecting the cricket eleven,then

$P=C(5,4)\times C(12,7)$

$\;\;\;=\large\frac{5!}{4!(5-4)!}\times\frac{ 12!}{7!(12-7)!}$

$\;\;\;=\large\frac{5\times 4!}{1!4!}\times\frac{ 12\times 11\times 10\times 9\times 8\times 7!}{7!5!}$

$\;\;\;=\large\frac{5}{1}\times \frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2\times 1}$

$\;\;\;=3960$

Hence (B) is the correct answer.

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