# Combinations: exam question distribution

• member 428835
In summary, the conversation discusses finding the number of ways to distribute questions among three students, with the constraint that each student must receive at least one question. The individual suggests using the stars and bars method, but realizes that it only works for identical questions. The solution key suggests using Stirling numbers, but the individual proposes a simpler approach of assigning the two easy questions first and then distributing the remaining questions among the students. This results in a total of ##2^{6-r}## ways for each value of r, leading to a final answer of ##\sum_{r=1}^6 2^{6-r}##.

#### member 428835

Homework Statement
Eight different exam questions are to be distributed among three
students, such that each student receives at least one question.
However, two of the questions are very easy and must be given to
different students. In how many ways can this be done?
Relevant Equations
stars and bars?
I think the idea is to first count the number of ways the questions are passed out without restrictions and then subtract the number of ways the two easy questions are together.

Call each student ##s_1,s_2,s_3##. We know their total number of questions is 8, so ##s_1+s_2+s_3 = 8 : s_1,s_2,s_3 > 0##, of course seeking only integer solutions. Thus we seek ##s_1+s_2+s_3 = 5## (each student gets at least one). I think stars and bars implies the unrestricted case is ##7C2##. Then if we lump two questions together (easy two) we have ##6C2##, thus the total of unrestricted is ##7C2-6C2 = 6##. This feels exceedingly low. What am I doing wrong?

Edit: just dawned on me this technique assumes all questions are identical. They are obviously not. So how to solve? My solution key states we use Stirling numbers. I guess I'll read about those.

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I would start by assigning the two easy questions. How many ways to do that?
One student has none yet. Choose r to give to that student, r=1..6. How many ways?
Dish out the remaining 6-r. How many ways?

member 428835
haruspex said:
I would start by assigning the two easy questions. How many ways to do that?
One student has none yet. Choose r to give to that student, r=1..6. How many ways?
Dish out the remaining 6-r. How many ways?
In order as you asked, ##3C2 \cdot 6Cr \cdot (6-r)C2##, though ##(6-r)C2## has to be wrong...maybe it's ##(6-r+2-1)! = (7-r)!##? So is the answer something like ##\sum_{r=1}^6 3C2 \cdot 6Cr \cdot (7-r)!##

Last edited by a moderator:
joshmccraney said:
In order as you asked, ##^3C_2 ##,
The two questions are different.
joshmccraney said:
##^{(6-r)}C_2##,
No, it's easier than that. Each of the 6-r can go to either of two students independently.

## 1. What is a combination in mathematics?

A combination is a way of selecting a group of items from a larger set without considering the order in which the items are selected. It is a mathematical concept used to calculate the number of possible outcomes when selecting a certain number of items from a larger set.

## 2. How is the number of combinations calculated?

The number of combinations is calculated using the formula nCr = n! / r!(n-r)!, where n represents the total number of items in the set and r represents the number of items being selected. This formula is also known as the combination formula or the "n choose r" formula.

## 3. What is the difference between a combination and a permutation?

The main difference between a combination and a permutation is that a combination does not consider the order of the items being selected, while a permutation does. In other words, a combination is a selection without replacement, while a permutation is a selection with replacement.

## 4. How can combinations be used in real-life situations?

Combinations can be used in various real-life situations, such as in probability and statistics, in genetics and biology, and in computer science and coding. For example, combinations can be used to calculate the number of possible outcomes when flipping a coin or rolling a dice, or to determine the number of possible genetic combinations in offspring.

## 5. How can combinations be applied to exam question distribution?

In exam question distribution, combinations can be used to ensure that the questions are distributed fairly and evenly across different topics and difficulty levels. By using combinations, the number of questions from each topic can be calculated based on the total number of questions and the desired distribution, ensuring that no topic is overrepresented or underrepresented.

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