$\begin{array}{1 1}(A)\;30\\(B)\;32\\(C)\;34\\(D)\;35\end{array} $

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Out of available nine courses,two are compulsory.Hence the student is free to select 3 courses out of 7 remaining courses.If P is the number of ways of selecting 3 courses out of 7 courses,then

$P=C(7,3)$

$\;\;\;=\large\frac{7!}{3!(7-3)!}$

$\;\;\;=\large\frac{7!}{3!4!}$

$\;\;\;=\large\frac{7\times 6\times 5\times 4!}{3\times 2\times 1\times 4!}$

$\;\;\;=35$ ways

Hence (D) is the correct answer.

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