1

A test consisting of $12$ questions is scored with $200$ points. How many ways can the $200$ points be assigned if each question must be worth at least $10$ points but no more than $25$ and the scores must be a multiple of $5$?

I'm thinking about a possible combination with repetition. $15$ points ($80/5$) out of $12$ questions. But I don't know how to apply the maximum restriction of $25$.$3$ to each question $\frac{25-10}{3}$.

But it may not be a combination with repetition either, as I interpret the statement as meaning that it doesn't matter if the first question is worth $30$ points and the second is worth $10$, which is the other way around. I understand that this specific example is a unique way, isn't it?

hajwpxsx
  • 13
  • 1
    You can start by allocating the required $10$ points to each question, which accounts for $120$ of the points, then dividing by $5$. Each question then can get $0$ to $3$ (more) points and the sum has to be $16$. Without the maximum of $3$ it would be a simple stars and bars problem. – Ross Millikan Nov 23 '20 at 19:11
  • @RossMillikan Yeah, the maximum of $3$ is what complicates it. – hajwpxsx Nov 23 '20 at 20:06
  • Once you've digested the problem, it can called counting the ${2,3,4,5}$-restricted compositions of $40$. It can be approached in various ways, as the two existing Answers may suggest to you. – hardmath Nov 23 '20 at 23:21

2 Answers2

1

Since each score must be a multiple of $5$, we might as well divide all of the point values by $5$ and have $12$ questions worth a total of $40$ points, each question being worth at least $2$ and at most $5$ points. If $p_k$ is the point value of the $k$-th question, we’re looking for the number of solutions to

$$\sum_{k=1}^{12}p_k=40\tag{1}$$

in integers $p_k$ satisfying the condition that $2\le p_k\le 5$ for $k=1,\ldots,12$. Let $x_k=p_k-2$ for $k=1,\ldots,12$; then the number of solutions to $(1)$ subject to the stated restriction is the same as the number of solutions to

$$\sum_{k=1}^{12}x_k=16$$

in non-negative integers $x_k$ satisfying the condition that $x_k\le 3$ for $k=1,\ldots,12$. If it weren’t for the upper limit on the numbers $x_k$, this would be a standard stars-and-bars problem, and there would be $\binom{16+12-1}{12-1}=\binom{27}{11}$ of them. Unfortunately, many of those solutions violate the upper limit on one or more of the numbers $x_k$, so $\binom{27}{11}$ is a significant overestimate. To correct this you will need to make an inclusion-exclusion calculation. My answer to this question includes such a calculation; try to use it as a model to complete the solution of your problem.

Brian M. Scott
  • 616,228
1

Since all scores are a multiple of $5$ we can divide by it, resulting in $40$ points total and question scores from $2$ to $5$ points. Since each question must be at least $2$ points, we can think of the questions as holding $24$ "base points", leaving the problem of how many ways there are to distribute $16$ "extra points" to the $12$ questions with no question having more than $3$ of them.

As a generating function problem, this is the $x^{16}$ coefficient of $(1+x+x^2+x^3)^{12}$. And the answer turns out to be $1501566$.

Parcly Taxel
  • 103,344