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I've been struggling with this question:

In how many ways can you grade 5 students so their average is 60? Note that each student's grade is an integer, no greater than 100 and no lower than 0.

I know that no more than 3 students can get a 100, but I don't see a way to count all the possible permutations for dividing 300 points into 5 baskets, so each of them is no more than 100 points.

I'd be grateful if you could hint me the right direction for solving this problem.

Ido
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  • What other constraints are there on the grades? I presume they need to be integers, but is this correct? – Cameron Buie Dec 25 '18 at 15:40
  • yes. i should have mentioned it. – Ido Dec 25 '18 at 15:41
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    You can adapt the technique presented in this answer, but with your numbers and constraints. – Rushabh Mehta Dec 25 '18 at 15:42
  • You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Shaun Dec 25 '18 at 15:42
  • By a similar argument, you can't have more than 2 students receiving 0's. Another way to solve such a problem would be with a computer and writing a script to count all the ways – WaveX Dec 25 '18 at 16:00
  • Possible duplicate of Combinatorial distribution – Ankit Kumar Dec 25 '18 at 16:18

2 Answers2

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We can also do it by generating function.

The generating function for the same is $(1+x+x^2+.. + x^{100})^{5}$ and what we want is the coefficient of $x^{300}$.

$(1+x+x^2+.. + x^{100})^{5} = (1-x^{101})^{5}.\frac{1}{(1-x)^{5}}$

$(1-x^{101})^{5}$ can be expressed $(1-{5\choose1}x^{101} + {5\choose2}x^{202}-{5\choose3}x^{303}..$

$$\frac{1}{(1-x)^{5}} = \sum_{0}^{\infty} {(n+5-1)\choose(5-1)}x^n$$

$\frac{1}{(1-x)^{5}} = \sum_{0}^{\infty} {(n+4)\choose(4)}x^n$

Multiplying these two expressions, you are looking to have n+4-r where r is the reducing number from the first expression. $r_i = 0, 101, 202$ for $r i = 0,1,2$ respectively

Thus coefficients are products of $(-1)^i({5\choose i}{(300+4-r_i)\choose 4}) x^{300}$

you get ${(300+4)\choose(4)}$ for the first term

you get the next one $- {5\choose1}{(300+4-101)\choose(4)}$

and the next one ${5\choose2}{(300+4-202)\choose(4)}$

Add these products such as below

$ {304\choose4}- {5\choose1}{203\choose4}+ {5\choose2}{102\choose(4)}=47952376 $

Satish Ramanathan
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I just made a spreadsheet. Start in row $101$, leaving all the ones above blank, and enter the numbers from $0$ through $300$ in column A. That will be the sum of grades. Now in column B enter $1$ in all the rows with numbers $0$ through $100$ in column A, which shows there is one way to give the first student any number of points from $0$ through $100$. Now in column C in line with $0$ points enter =SUM(Left up 100:Left), where you substitute the cell references for the directions, which says the number of ways to get a sum of $0$ for the first two students is the sum of the $101$ lines ending in this one. Copy down and right for $300$ points and $5$ students. I find the total to be $45235674$

The generating function approach is that the function for one student is $1+x+x^2+\ldots x^{100}=\frac {1-x^{101}}{1-x}$ so the function for five students is $$\left(\frac {1-x^{101}}{1-x}\right)^5$$ and you want the coefficient of $x^{300}$ of this. I don't have an easy way to evaluate that, but I think Mathematica can do so.

Ross Millikan
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