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I'm having trouble with this problem. Any help you can offer would be appreciated.

Abigail, Bruno, Candy, Dave, Eliza, and Fred are the 6 students in Mr. Smith's beginning computer class. It is known that Mr. Smith only uses 3 grades when he grades a test-A, B, or C. All six students took the first test, Abigail received and A on the test, there were exactly twice as many B's as A's on the test, and it is not known whether or not any students received a C on the test. Find the number of different ways for the grades of the 6 students on the first test. (Note: An A for Bruno and a B for Candy is different than an A for Candy, and a B for Bruno, for example.)

Thanks a lot.

-Isaac

We first notice that Abigail received an A and so the problem is really how many ways that the A, B and Cs can be distributed among the remaining 5 students. It is also very important that there are exactly 2 Bs for every A.

So to start, if only Abigail gets an A, there must be exactly 2 Bs among the 5 other students and the rest Cs. This is the same as asking how many ways can 2 things (the Bs) be distributed among 5 objects (the students). Note that the Bs can be interchanged among the students but the Bs and Cs cannot so that BB is one outcome while CB and BC are distinct. The number of ways then is given by the combination formula

5! / 2!3! = 10

You can prove this to yourself by writing out the possibilities

BBCCC

BCBCC

BCCBC

BCCCB

CBBCC

CBCBC

CBCCB

CCBBC

CCBCB

CCCBB

= 10 combinations.

If Abigail is not the only one to get an A, then we can assign an A to one of the other students, but importantly, we have to include a total of 4 Bs as well. Thus, we need to distribute 1 A among the 5 students, so like above, we have

5! / 1!4! = 5

Again, there aren't so many numbers that we can't simply write them out (1 A and 4 Bs among 5 students)

ABBBB

BABBB

BBABB

BBBAB

BBBBA

= 5 combinations.

Now we can consider giving 3 As to the students but we immediately run into the problem of 3As -> 6 Bs and we only have 6 students, so this situation is not possible. So the total numbers of ways is 15.

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