- #1

#### 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?

I appreciate your help.

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.

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?

I appreciate your help.

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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