Combination & Distribution: Solving for Musical Instrument Assignments

[Glen Maverick]

Combination & distribution

Homework Statement

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.

Homework Equations

Combination equation

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.

 

[tiny-tim]

Hi Glen!

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.
…
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.

Yes.

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]

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! ​

 

[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?

 

[tiny-tim]

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

That's it!

And Have I got right in the event b?

No, I don't really follow what you've done there.

 

[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).

 

[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.

 

[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!