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Maths-

General

Easy

Question

A man goes in for an examination in which there are four papers with a maximum of 'm' marks for each paper then the number of ways of getting 2m marks is

$\frac{(m - 1) (2 m^{2} + 4 m + 3)}{3}$
$\frac{(m + 1) (2 m^{2} + 4 m + 3)}{3}$
$\frac{(m + 1) (2 m^{2} - 4 m + 3)}{3}$
None of these

The correct answer is: $fraction numerator left parenthesis m plus 1 right parenthesis left parenthesis 2 m to the power of 2 end exponent plus 4 m plus 3 right parenthesis over denominator 3 end fraction$

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