Question 90

Mathematics Permutation and Combination Medium

A person goes in 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 2 m marks is

(A) \( \frac{1}{3}(m+1)\left(2 m^{2}+4 m+1\right) \)
(B) \( \frac{1}{3}(m+1)\left(2 m^{2}+4 m+2\right) \)
(C) \( \frac{1}{3}(m+1)\left(2 m^{2}+4 m+3\right) \)
(D) \( { }^{2 m+3} G_{3} \)
View Dynamic Solution & Explanation
Correct Solution: Option C

Step-by-step Solution:

\( \text{The required number} = \text{Coeff of } x^{2m} \text{ in } (1 - x^m + 1)^4 (1 - x)^{-4} \)
\( = \text{Coeff of } x^{2m} \text{ in } (x^0 + x^1 + \dots + x^m)^4 \)
\( = \frac{(2m + 1)(2m + 2)(2m + 3)}{6} - \frac{4m(m + 1)(m + 2)}{6} \)
\( = \frac{(m + 1)(2m^2 + 4m + 3)}{3} \)