(More) Combinatorics

Hello again mathhelp forum,

I've come across a problem in an exercise. It goes like this:

A class consists of 50 students. They will be graded after a test, and each student can get the grade A, B or C. It is decided beforehand that 10 students will be given A, 20 B and 20 C.

Furthermore, there are three students in this class named Lisa, Kalle and Gunbert.

In how many ways can the class be graded so that none of these three get a C?

If you need any additional information about the exercise please let me know.

cloudhorn said:

A class consists of 50 students. They will be graded after a test, and each student can get the grade A, B or C. It is decided beforehand that 10 students will be given A, 20 B and 20 C.

Furthermore, there are three students in this class named Lisa, Kalle and Gunbert.

In how many ways can the class be graded so that none of these three get a C?

Click to expand...

What are your thoughts? What have you tried so far? Where are you stuck?

Please be complete. Thank you!

I've tried thinking that there are 3 students who can't get grade C, so all in all there are 47 students who can. And because there is a total of 20 C's there are (47 choose 20) possible combinations for grade C (supposing we've started with this grade).

Now there are 27 + 3 = 30 students who have not yet been given their grades, and they can get A or B. Since there are 20 B's, I'm thinking there are now (30 choose 20) possible combinations for grade B, and then finally 1 combination left for grade A.

Where I'm stuck is deciding whether that's it, or if I should also count the amount of combinations that Kalle, Lisa and Gunbert (the 3 students) can get the grades A and B. Since there are 2 grades to be distributed among 3 students I get this to be 3!.

However, my teacher gave me a bunch of questionmarks when I handed in that answer (which in the end became (47 choose 20) * (30 choose 20) * 3!), so there seems to be an error somewhere.

Does any of this seem to make sense?

cloudhorn said:

Hello again mathhelp forum,

I've come across a problem in an exercise. It goes like this:

A class consists of 50 students. They will be graded after a test, and each student can get the grade A, B or C. It is decided beforehand that 10 students will be given A, 20 B and 20 C.

Click to expand...

It was decided beforehand that no one will get a D or an F? Grade inflation!

There are 50(49)(47)...(41)= 50!/40! ways of choosing 10 students to get an A. Once that is done there are 40 students left and there are 40(39)(38)...(21)= 40!/20! ways of choosing 20 of them to get a B. That's leaves the last 20 to get a C. There are (50!/40!)(40!/20!)= 50!/20! ways of assigning such grades.

Furthermore, there are three students in this class named Lisa, Kalle and Gunbert.

In how many ways can the class be graded so that none of these three get a C?

If you need any additional information about the exercise please let me know.

Click to expand...

Give "Lisa" a C to begin with and determine how many ways there are to give 10 A's, 20 B's, and 19 C's to the other 49 students. Do the same with Kalle and Gunbert (obviously getting the same result). Add them together (or multiply the first answer by 3) to find the number of ways of assigning grades that gives a C to at least one of the three. The number of ways of assigning grades so that Lisa, Kalle, and Gunbert do NOT get C is the original number of ways of assigning grades minus the number of ways such that at least one does get a C.

HallsofIvy said:

There are 50(49)(47)...(41)= 50!/40! ways of choosing 10 students to get an A. Once that is done there are 40 students left and there are 40(39)(38)...(21)= 40!/20! ways of choosing 20 of them to get a B. That's leaves the last 20 to get a C. There are (50!/40!)(40!/20!)= 50!/20! ways of assigning such grades.

Click to expand...

I think you're forgetting that the order in which the students are assigned their grades doesn't matter. I argue that it doesn't matter if, say a student from the class named Kevin, is the first one to get an A or the second one, and so on. To make up for this it's possible to the use the binomial coefficient (50 choose 10) = 50!/(10!*40!).

Your argument on how to solve the problem is interesting though and I'll have to check it out.