Question

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 2m marks is

• A. $\frac{1}{3}(m+1)(2m^2+4m+1)$
• B. $\frac{1}{3}(m+1)(2m^2+4m+2)$
• C. $\frac{1}{3}(m+1)(2m^2+4m+3)$
• D. $^{2m+3}C_3$

Answer 1

Answer: (C)

$\frac{1}{3}(m+1)(2m^2+4m+3)$

Answer 2

The correct answer is(C): $\frac{1}{3}(m+1)(2m^2+4m+3)$