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In an examination a student has to answer 4 questions out of 5 questions;questions 1 and 2 are compulsory.Determine the number of ways in which the student can make choice.

$\begin{array}{1 1}(A)\;2\\(B)\;3\\(C)\;4\\(D)\;5\end{array} $

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Total no of questions =5
Total no of questions to be answered =4
Compulsory questions =1 and 2
Hence the remaining questions are
Hence the no of ways a student can make choice =$3C_2$
$\Rightarrow \large\frac{3!}{2!\times 1!}$
$\Rightarrow \large\frac{3\times 2!}{2!}$
$\Rightarrow 3$
Hence (B) is the correct answer.
answered May 15, 2014 by sreemathi.v

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