Best way to distribute exam questions to students for a fair exam

[member 428835]

Homework Statement

8 different exam questions are to be distributed among 3
students, such that each student receives at least one question.
However, 2 of the questions are very easy and must be given to
different students. In how many ways can this be done?

Relevant Equations

Nothing comes to mind

Without the extra restriction, the number of ways to do this is $3!S ( 8, 3 )$: The Stirling number counts ways to distribute the different questions into three non-empty sets and the 3! accounts for the different ways to distribute the sets among the students.

To construct the forbidden configurations, we can combine the easy questions and distribute the now 7 questions among the students. This can be done in $3!S ( 7, 3 )$ ways.

The final answer is therefore $3!S ( 8, 3 ) − 3!S ( 7, 3 )$. Now when I plug this into Mathematica via the StirlingS2 function I get 3990, which must be way too large a number. Does anyone know what the number should be?

 

[andrewkirk]

Divide it into steps, and calc the answer as the product of the number of different ways to do each step. Assume the questions are numbered 1-8 and questions 1-2 are the easy ones.

Step 1: Choose a student to do Q1
Step 2: Choose one of the other two students to do Q2
Step 3: Choose one of questions 3-8 to give to the remaining student, so that every student has one question.
Step 4: Allocate the remaining 5 questions amongst the three students without restriction.

I get 8748

 

[member 428835]

Divide it into steps, and calc the answer as the product of the number of different ways to do each step.

Step 1: Give the first easy question to a student
Step 2: Give the second easy question to one of he other two students.
Step 3: Give one of the remaining 5 questions to the remaining student, so that every student has one question.
Step 4: Allocate the remaining 5 questions amongst the three students without restriction.

I get 8748

Isn't it simpler to consider all possible combinations without limitations and then subtract the violation, which is the case where both easy questions go to the same student?

 

[andrewkirk]

I find my method very simple. For a start one doesn't have to worry about Stirling numbers and can do the calc in one's head.
Try it and see.

 

[member 428835]

I find my method very simple. For a start one doesn't have to worry about Stirling numbers and can do the calc in one's head.
Try it and see.

But is the correct answer yours or mine? Tough for me to tell

 

[member 428835]

I find my method very simple. For a start one doesn't have to worry about Stirling numbers and can do the calc in one's head.
Try it and see.

So I'm getting: $3C1*2C1*5C1*3^5 = 7290$

 

[andrewkirk]

The third factor should be 6C1 not 5C1.

 

[Orodruin]

Just to note that with the restrictions described, this exam can become very very unfair.

Student 1: Easy problem.
Student 2: Easy problem.
Student 3: 6 hard problems.

 

[vela]

Divide it into steps, and calc the answer as the product of the number of different ways to do each step. Assume the questions are numbered 1-8 and questions 1-2 are the easy ones.

Step 1: Choose a student to do Q1
Step 2: Choose one of the other two students to do Q2
Step 3: Choose one of questions 3-8 to give to the remaining student, so that every student has one question.
Step 4: Allocate the remaining 5 questions amongst the three students without restriction.

I get 8748

It does seem a bit strange that using the Stirling number with no restrictions multiplied by the number of ways to permute the partition among the students gives a smaller number (5796) than your answer.

 

[vela]

The final answer is therefore $3!S ( 8, 3 ) − 3!S ( 7, 3 )$. Now when I plug this into Mathematica via the StirlingS2 function I get 3990, which must be way too large a number. Does anyone know what the number should be?

I think your answer is right, but I don't see what's wrong with @andrewkirk's calculation.

I had Mathematica generate all the partitions of eight questions into three subsets, which results in 966 possibilities. Then I had it count all of those partitions where questions 1 and 2 were assigned to the same student. There were 301 of those, leaving 665 good partitions. That number multiplied by 3! gives 3990.

 

[vela]

I think I see what's wrong with @andrewkirk's approach. Say Alice gets Q1, Bob gets Q2, and Carlos gets Q3 and then the rest are randomly assigned such that Carlos ends up with Q3 and Q4. Andrew's method would count that possibility as separate from Carlos getting Q4 initially and Q3 later while Alice and Bob receive the same questions as before.

 

[member 428835]

I think I see what's wrong with @andrewkirk's approach. Say Alice gets Q1, Bob gets Q2, and Carlos gets Q3 and then the rest are randomly assigned such that Carlos ends up with Q3 and Q4. Andrew's method would count that possibility as separate from Carlos getting Q4 initially and Q3 later while Alice and Bob receive the same questions as before.

Okay, this explanation makes sense to me, thanks for clarifying.