Word Problem Involving Permutations/Combinations

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mikie

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I'mnot sure if this is the right place to post, but i have a question that i don't know where to begin with.


Suppose 10 indistinguishable dimes and 6 indistinguishable quarters are to be distributed among 6 people: A, B, C, D, E and F. How many ways can this be done if it is possible for any person to receive no coins?
 
The problem is that i'm not sure where to begin. I think it's a combinations problem because the coins have no specific, order ( as they are indistinguishable). but even then, i'm not sure how to answer the question.

My thought was to first find out, how many ways the quarters can be distributed among the 6 people, then find out how many ways the dimes can be distrubted among the 6 people, and then multiply the two together. But i'm not sure how, even if it is a reasonable method of solving the porblem., to do it.

I know that the answer is something like 1 300 000. Though i really don't know how to get there

if some one can give me a hand, that would be much appriciated. thanks!

mike
 
You are absolutely correct, “how many ways the quarters can be distributed among the 6 people, then find out how many ways the dimes can be distrubted among the 6 people, and then multiply the two together.”

Combin(10+6−1,10)*Combin(6+6−1,6)=1387386.
 
aw, hey, thanks for such a quick response.

:oops: (it's my first time on the forums)

would you be able to explain the arithmetic. (i.e. why is it 10 + 6 -1) and (6+6 - 1)
 
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wow, this one is definitely a head scratcher for me.

i understand the basic concept of explanation, though i get completely lost with the stated equation. m+n-1 factorial / m factorial x n-1 factorial.

can you sort of elaborate on this...if at all possible? and show how you got from that equation to m +n -1



I'm terribly sorry and appreciate your time and efforts, in putting up with my lack of understanding.
 
Well maybe it is just best for now for you to learn the formula.
The number of ways to put k identical objects into n different cells is Combin(k+n−1,k).

So to give 10 dimes to six people, the dimes are identical objects and the people are different cells: Combin(10+6−1,10).

The number of non-negative integer solutions to a+b+c+d=15 is Combin(15+4−1,15).
We put 15 identical ones into 4 different variables.
 
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