# Combinations: exam question distribution

Gold Member
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.

Last edited:

Homework Helper
Gold Member
2022 Award
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?

joshmccraney
Gold Member
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: