# Combination problem help?

1. Feb 1, 2010

### Glen Maverick

Combination & distribution

1. The problem statement, all variables and given/known data

Six musical instruments are available for loan. Assuming all are loaned, in how many different ways can these be assigned to the four musicians in the graduate music ensemble such that each instrument is loaned to one musician, each musician gets at least one instrument and no musician has more than three instruments on loan? Give answer as a whole number.

2. Relevant equations

Combination equation

3. The attempt at a solution

At first, I tried to solve this problem based on other similar problem's solution.
Here's how I tried:
There are two possibilities:
a - a student having 3 instruments and others have only one.
b - two students having two instruments and the rest have only one.

event a:[C(4,1) x C(6,3) x C(3,1) x C(3,1) x C(3,1) divided by 46]
event b:[C(4,1) x C(6,2) x C(4,2) x C(2,1) x C(1,1) divided by 46]

And Since two events I showed above are mutually exclusive, I can do like this: a+b.
Is this the right procedure? Please check for me.

Last edited: Feb 1, 2010
2. Feb 2, 2010

### tiny-tim

Hi Glen!
Yes.
No.

event a is ok up to C(4,1) x C(6,3) …

that's the number of ways of choosing the student who gets 3, times the number of ways of giving him 3.

But now you have only 3 left, and you need the number of ways of giving them to one each of 3, and that is not C(3,1) x C(3,1) x C(3,1), is it?

Try again!

3. Feb 2, 2010

### Glen Maverick

Oh! Now I get it. C(4,1) x C(6,3) x C(3,1) x C(3,1) x C(3,1) divided by 4^6 is not right, because if one of the student picks 1 among three, the leftover would be two.
So. C(4,1) x C(6,3) x C(3,1) x C(2,1) x C(1,1) divided by 4^6
And Have I got right in the event b?

4. Feb 2, 2010

### tiny-tim

That's it!
No, I don't really follow what you've done there.

5. Feb 2, 2010

### Glen Maverick

Thank you so much for checking for me!

b - two students having two instruments and the rest have only one.
event b:[C(4,1) x C(6,2) x C(4,2) x C(2,1) x C(1,1) divided by 4^6]
First student is chosen among 4, and he or she chooses two instuments from 6. Second student also choses 2, but among 4, since two is already taken by the first student. And what left is 2 instruments. And two students are left. They take one each from the leftover. That is why I did C(6,2) x C(4,2) x C(2,1) x C(1,1).

6. Feb 3, 2010

### tiny-tim

(just got up :zzz: …)

You haven't described how the second student is chosen.

You need to choose two students to get two instruments, and then you need to choose four instruments to go to those two students. Then you need to count the ways of dividing those four instruments among those two, and also the other two instruments among the other two.

7. Feb 6, 2010

### Glen Maverick

Now I understand it! I was quite troubling with this kind of problems. Sorry for the late reply... Thank you so much for helping me. :) Have a good day!