A person takes an examination consisting of f | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The number of ways to get a total of $2m$ marks in $4$ papers,where each paper has a maximum of $m$ marks,is the coefficient of $x^{2m}$ in the ex

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The number of ways to get a total of $2m$ marks in $4$ papers,where each paper has a maximum of $m$ marks,is the coefficient of $x^{2m}$ in the expansion of $(x^0 + x^1 + \dots + x^m)^4$.This is equal to the coefficient of $x^{2m}$ in $\left(\frac{1 - x^{m+1}}{1 - x}ight)^4$.$= 	ext{coefficient of } x^{2m} 	ext{ in } (1 - x^{m+1})^4 (1 - x)^{-4}$.$= 	ext{coefficient of } x^{2m} 	ext{ in } (1 - 4x^{m+1} + 6x^{2m+2} - \dots) \left(\sum_{r=0}^{\infty} \binom{r+3}{3} x^right)$.$= \binom{2m+3}{3} - 4 \binom{m+3-1}{3} = \binom{2m+3}{3} - 4 \binom{m+2}{3}$.$= \frac{(2m+3)(2m+2)(2m+1)}{6} - 4 \frac{(m+2)(m+1)m}{6}$.$= \frac{(m+1)}{6} [ (2m+3)(2)(2m+1) - 4m(m+2) ]$.$= \frac{(m+1)}{6} [ 2(4m^2 + 8m + 3) - 4m^2 - 8m ]$.$= \frac{(m+1)}{6} [ 8m^2 + 16m + 6 - 4m^2 - 8m ]$.$= \frac{(m+1)(4m^2 + 8m + 6)}{6} = \frac{(m+1)(2m^2 + 4m + 3)}{3}$.

$A$ person takes an examination consisting of four papers,each with a maximum of $m$ marks. The number of ways in which one can obtain a total of $2m$ marks is

Similar Questions

The greatest coefficient in the expansion of $(1 + x)^{2n + 1}$ is

Show that the coefficient of the middle term in the expansion of $(1+x)^{2n}$ is equal to the sum of the coefficients of the two middle terms in the expansion of $(1+x)^{2n-1}$.

If the coefficients of $r$ th and $(r+1)$ th terms in the expansion of $(3+7x)^{29}$ are equal,then $r$ is equal to

If the $4^{\text{th}}$ term in the expansion of $\left(\frac{x}{2}-\frac{2y}{3}\right)^6$ is $-20$,then $xy=$

The greatest term in the expansion of $\sqrt{3} \left( 1 + \frac{1}{\sqrt{3}} \right)^{20}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module