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) ${}^{2m+3}{{C}_{3}}$
(b) $\dfrac{1}{3}\left( m+1 \right)\left( 2{{m}^{2}}+4m+1 \right)$
(c) $\dfrac{1}{3}\left( m+1 \right)\left( 2{{m}^{2}}+4m+3 \right)$
(d) ${}^{2m-3}{{C}_{3}}$

Answer 1

We start solving the problem by assigning the variables for the marks scored in each exam. We then recall the fact that the number of ways in which ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......+{{a}_{r}}=n$, with 0 allowed for ${{a}_{i}}'s$ is ${}^{n+r-1}{{C}_{r-1}}$ to get the number of ways to score 2m marks. We then find the number of ways that the marks in a single exam exceeds m marks which will be subtracted from the number of ways to get the required answer.

Complete step by step answer:
According to the problem, we are given that a person goes in for an examination in which there are four papers with a maximum of m marks from each paper. We need to find the number of ways in which one can get 2m marks.
Let us assume that the marks scored in each exam are x, y, z and w.
So, we need to find the number of ways that $x+y+z+w=2m$ ---(1) and $0\le x,y,z,w\le m$.
We know that the number of ways in which ${{a}_{1}}+{{a}_{2}}+{{a}_{3}}+......+{{a}_{r}}=n$, with 0 allowed for ${{a}_{i}}'s$ is ${}^{n+r-1}{{C}_{r-1}}$.
So, we get the number of ways as ${}^{2m+4-1}{{C}_{4-1}}={}^{2m+3}{{C}_{3}}$ ---(2). But these ways include the ways that $x,y,z,w\ge m-1$.
Let us assume $t=x-\left( m+1 \right)$ which gives $t\ge 0$. Let us substitute this in equation (1).
$\Rightarrow t+m+1+y+z+w=2m$.
$\Rightarrow t+y+z+w=m-1$.
So, the number of ways will be ${}^{m-1+4-1}{{C}_{4-1}}={}^{m+2}{{C}_{3}}$ ---(3).
Let us assume $s=y-\left( m+1 \right)$ which gives $t\ge 0$. Let us substitute this in equation (1).
$\Rightarrow x+s+m+1+z+w=2m$.
$\Rightarrow x+s+z+w=m-1$.
So, the number of ways will be ${}^{m-1+4-1}{{C}_{4-1}}={}^{m+2}{{C}_{3}}$ ---(4).
Let us assume $u=z-\left( m+1 \right)$ which gives $t\ge 0$. Let us substitute this in equation (1).
$\Rightarrow x+y+u+m+1+w=2m$.
$\Rightarrow x+y+u+w=m-1$.
So, the number of ways will be ${}^{m-1+4-1}{{C}_{4-1}}={}^{m+2}{{C}_{3}}$ ---(5).
Let us assume $v=w-\left( m+1 \right)$ which gives $t\ge 0$. Let us substitute this in equation (1).
$\Rightarrow x+y+z+v+m+1=2m$.
$\Rightarrow x+y+z+v=m-1$.
So, the number of ways will be ${}^{m-1+4-1}{{C}_{4-1}}={}^{m+2}{{C}_{3}}$ ---(6).
Let us subtract equations (3), (4), (5), (6) from equation (2).
So, we get the required number of ways as ${}^{2m+3}{{C}_{3}}-{}^{m+2}{{C}_{3}}-{}^{m+2}{{C}_{3}}-{}^{m+2}{{C}_{3}}={}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}$.
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.
$\Rightarrow {}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}=\dfrac{\left( 2m+3 \right)!}{3!2m!}-4\times \dfrac{\left( m+2 \right)!}{3!\left( m-1 \right)!}$.
$\Rightarrow {}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}=\left( \dfrac{\left( 2m+3 \right)\times \left( 2m+2 \right)\times \left( 2m+1 \right)}{3\times 2\times 1} \right)-4\times \left( \dfrac{\left( m+2 \right)\times \left( m+1 \right)\times \left( m \right)}{3\times 2\times 1} \right)$.
$\Rightarrow {}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}=\left( m+1 \right)\times \left( \left( \dfrac{4{{m}^{2}}+8m+3}{3} \right)-\left( \dfrac{2\times \left( {{m}^{2}}+2m \right)}{3} \right) \right)$.
$\Rightarrow {}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}=\left( m+1 \right)\times \left( \left( \dfrac{4{{m}^{2}}+8m+3}{3} \right)-\left( \dfrac{2{{m}^{2}}+4m}{3} \right) \right)$.
$\Rightarrow {}^{2m+3}{{C}_{3}}-4{}^{m+2}{{C}_{3}}=\dfrac{1}{3}\left( m+1 \right)\left( 2{{m}^{2}}+4m+3 \right)$.
So, we have found the required number of ways as $\dfrac{1}{3}\left( m+1 \right)\left( 2{{m}^{2}}+4m+3 \right)$.

So, the correct answer is “Option c”.

Note: We can see that the given problem contains a huge amount of calculations so, we need to perform each step carefully in order to avoid confusion and calculation mistakes. We should not forget to subtract the number of ways in which the marks in a single exam is greater than m, which is the common mistake done by students. Similarly, we can expect problems to find the total number of ways to score 3m marks together in all exams.