Help me solve this Finite Math problem

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12 years ago 01/30/2012 1:57pm EST

My professor gave us the following question:

There are 15 limousines. If the people need 4 of them to go to party A, 3 to party B, 3 to party C, and 5 to party D, how many different selections of limousines can be made?

I believe it's a combination problem, but I suppose it also could be a permutation. I'm trying to understand how to solve this problem.

I already know that P(n,r)-->(n!)/(n-r)! and c(n,r)-->(n!)/(n-r)!(r)! ...But I don't know how to use them in this particular case.

Any help is appreciated.

12 years ago 01/30/2012 2:03pm EST

re: Gate-a-raid

Gate-a-raid wrote:
My professor gave us the following question:
There are 15 limousines. If the people need 4 of them to go to party A, 3 to party B, 3 to party C, and 5 to party D, how many different selections of limousines can be made?
I believe it's a combination problem, but I suppose it also could be a permutation. I'm trying to understand how to solve this problem.
I already know that P(n,r)-->(n!)/(n-r)! and c(n,r)-->(n!)/(n-r)!(r)! ...But I don't know how to use them in this particular case.
Any help is appreciated.

Couldn't they just use one?

12 years ago 01/30/2012 2:06pm EST

re: Randy Oldman

Randy Oldman wrote:
Couldn't they just use one?

It would have made my life a lot easier if they had...

12 years ago 01/30/2012 2:37pm EST

re: Gate-a-raid

Without putting too much thought into it:

C(15,4) * C(11,3) * C(8,3) * C(5,5)

12 years ago 01/30/2012 2:44pm EST

re: Gate-a-raid

c(15,4) * c(11,3) * c(8,3)

12 years ago 01/30/2012 2:57pm EST

re: Intelligenter than thou

In general, if you have X objects and you want to put them in groups of g1, g2, g3, ..., gN, then the total number of combinations would be X!/(g1! * g2! * g3! * ... * gN!)

12 years ago 01/30/2012 3:22pm EST

re: Gate-a-raid

Dude, math, by definition, is infinite. Your problem doesn't even exist.

Next.

12 years ago 01/30/2012 4:30pm EST

re: Gate-a-raid

Can limousines be reused? If so then it would be

c(15,4)*c(15,3)*c(15,3)*c(15,5)

If they can't then:

c(15,4)*c(11,3)*c(8,3)*c(5,5)

this reduces to:

15!/(4!3!3!5!) just as the people above said.

12 years ago 01/30/2012 6:35pm EST

re: fiftypence

So I was supposed to multiply the C(15,4)xC(11,3)xC(8,3)? (Saying the c(5,5)=1.) I wouldn't add them together?

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