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

^{2 m + 3} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^{2 m - 3} C_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

A person goes in for an examination in which there are four papers with a maximum of 20 marks from each paper. The number of ways in which one can get 40 marks is_______.

Q. A man goes in for an examination in which there are

4

papers with a maximum of

10

marks for each paper. The number of ways of getting

20

marks on the whole is

Q. In an examination, there are

4

papers with a maximum of 'n' marks for each paper. Then the number of ways in which can one get

50 %

marks is?

Q. In an examination the maximum marks for each of the three papers are

50

each. maximum marks for the fourth paper

100.

The number of ways in which the candidate can score

60

% marks in aggregate is

Q. In an examination, the maximum marks for each of three papers are

50

each. Maximum marks for the

4^{t h}

paper is

100

. Find the number of ways in which the candidate can score

60

% marks in the aggregate.

Join BYJU'S Learning Program

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

The correct option is C 13(m+1)(2m2+4m+3)The required number =Coeff of x2m in (1−xm+1)4(1−x)−4= Coeff of x2m in (x0+x1+…+xm)4 =(2m+1)(2m+2)(2m+3)6 −4m(m+1)(m+2)6 =(m+1)(2m2+4m+3)3

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