Q. A person goes for an examination in which there are four papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks, is
A) $2m + 1$
B) $\frac{1}{3}(m + 1)(2m^{2} + m + 1)$
C) $\frac{1}{3}(m + 1)(2m^{2} + 4m + 3)$
D) None of the above
Explanation
(c)
Required number of ways
=coefficient of $x^{2m}$in $\left( x^{0} + x^{1} + ... + x^{m} \right)^{4}$
=coefficient of $x^{2m}$ in $\left( \frac{1 - x^{m + 1}}{1 - x} \right)^{4}$
=coefficient of $x^{2m}$ in $\left( 1 - 4x^{m + 1} + 6x^{2m + 2} + ... \right)(1 - x)^{- 4}$
$=^{2m + 3}C_{2m} - 4^{m + 2}C_{m - 1}$
$= \frac{(2m + 1)(2m + 2)(2m + 3)}{6} - \frac{4m(m + 1)(m + 2)}{6}$
$= \frac{(m + 1)(2m^{2} + 4m + 3)}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.