Combination of problem having identical objects without using generating function approach

Question

A person goes in for an examination in which there are $4$ papers with a maximum of $m$ marks from each paper. The number of ways in which one can get $2m$ marks is

My Attempt

Number of ways of partitioning $n$ identical objects into $r$ distinct groups, if empty groups are allowed is:${}^{n+r-1}C_{r-1}$.

Here, $n=2m$ and $r=4$

Req. number of ways = ${}^{n+r-1}C_{r-1}-$[#{more than m marks for paper 1}+#{more than m marks for paper 2}+#{more than m marks for paper 3}+#{more than m marks for paper 4}]=$${}^{2m+4-1}C_{4-1}-[{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}+{}^{m+4-1}C_{4-1}]={}^{2m+4-1}C_{4-1}-4.{}^{m+4-1}C_{4-1}\\ ={}^{2m+3}C_{3}-4.{}^{m+3}C_{3}=\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4\frac{(m+3)(m+2)(m+1)}{3.2}\\ =\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12]=\frac{(m+1)(2m^2-2m-9)}{3} $$

But my reference gives the solution $\dfrac{(m+1)(2m^2+4m+3)}{3}$, so what is going wrong here ?

Note: I am trying to solve this problem without using the generating function approach

Answer 1

$\#$ of ways $n$ identical objects can be divided into $r$ distinct groups if empty groups are allowed is:${}^{n+r-1}C_{r-1}$

Req. # of ways=$${}^{2m+4-1}C_{4-1}-\Big[\text{#{> m for paper 1}+#{> m for paper 2}+#{> m for paper 3}+#{> m for paper 4}}\Big]\\+\Big[\text{#{strictly m for paper 1}+#{strictly m for paper 2}+#{strictly m for paper 3}+#{strictly m for paper 4}}\Big]\\ ={}^{2m+4-1}C_{4-1}-4\Big[{}^{m+4-1}C_{4-1}\Big]+4\Big[{}^{m+3-1}C_{3-1}\Big]\\ ={}^{2m+3}C_{3}-4\Big[{}^{m+3}C_{3}\Big]+4\Big[{}^{m+2}C_{2}\Big]\\ =\frac{(2m+3)(2m+2)(2m+1)}{3.2}-4.\frac{(m+3)(m+2)(m+1)}{3.2}+4.\frac{(m+2)(m+1)}{2}\\ =\frac{m+1}{3}.[4m^2+8m+3-2m^2-10m-12+6m+12]=\frac{m+1}{3}.[2m^2+4m+3] $$

Answer 2

You should consider $\#\{\text{strictly more than $m$ marks for paper $j$}\}$. So we start with $m+1$ marks in paper $j$ and then distribute the remaining $m-1$ marks among the 4 papers. Thus we get $$\frac{(2m+3)(2m+2)(2m+1)}{3\cdot2}-4\frac{(m+2)(m+1)m}{3\cdot2}$$ $$=\frac{m+1}{3}(4m^2+8m+3-2m^2-4m)$$ $$=\frac{(m+1)(2m^2+4m+3)}{3}$$

Answer 3

We have to count the number of solutions of $$x_1+x_2+x_3+x_4=2m$$ where $0\leq x_i\leq m$ are integers for $i=1,2,3,4$. By Stars and Bars, if we consider just the condition $x_i\geq 0$ we have $\binom{2m+3}{3}$ solutions. On the other hand, we can have at most one variable $x_i\geq m+1$. If $i=1$, we have to count the number of solutions of $$y_1+x_2+x_3+x_4=2m-(m+1)=m-1$$ where $y_1:=x_1-(m+1)\geq 0$, $x_i\geq 0$ for $i=2,3,4$. Such solutions, again by Stars and Bars, are $\binom{(m-1)+3}{3}$. By symmetry, the same number turns out for $i=2,3,4$. Hence the answer is $$\binom{2m+3}{3}-4\binom{m+2}{3}=\frac{(m+1)(2m^2+4m+3)}{3}.$$