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Topic: balls in bags (Read 1146 times) |
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inexorable
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balls in bags
« on: Jan 7th, 2006, 11:47am » |
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m identical balls are to be placed in n distinct bags. You are given that m>=kn, where k is a natural number >=1. In how many ways can the balls be placed in the bags if each bag must contain atleast k balls?
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towr
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Re: balls in bags
« Reply #1 on: Jan 8th, 2006, 7:46am » |
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choose(m, kn) * choose(m-kn+n-1, m-kn)
randomly pick the kn balls you have to put in the n bags
then distribute the rest randomly.
I'm not entirely convident though 
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Grimbal
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Re: balls in bags
« Reply #2 on: Jan 8th, 2006, 2:39pm » |
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I'd say put k balls in each bag and count in how many ways you can arrange the m-kn remaining.
The ways to put m-kn balls in n bags is Choose((m-kn)+n,n).
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towr
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Re: balls in bags
« Reply #3 on: Jan 8th, 2006, 11:58pm » |
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on Jan 8th, 2006, 2:39pm, Grimbal wrote:The ways to put m-kn balls in n bags is Choose((m-kn)+n,n).
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I don't think so. n bags, means n-1 seperators.
And the number of ways to put 3 balls in one bag is 1, not 4.
And if you have to put two balls in two bags, one in each, you can do so in two ways. Put one in bag one first, or in bag two first, and then one in the other.
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| « Last Edit: Jan 8th, 2006, 11:59pm by towr » |
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Grimbal
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Re: balls in bags
« Reply #4 on: Jan 9th, 2006, 12:54am » |
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Uh... well, yes, there must be a -1 somewhere.
Choose(m-kn+n-1,n-1) maybe?
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towr
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Re: balls in bags
« Reply #5 on: Jan 9th, 2006, 11:24am » |
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That solves half the problem.
If m=k*n, as an obvious case, you'd always get 1, while I'd say it's clearly not.
And I suppose it also discredits my first solution.
for k=1 if m=n, there are m! ways to fill the bags
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Grimbal
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Re: balls in bags
« Reply #7 on: Jan 10th, 2006, 12:28am » |
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I'm not sure it is about the process of putting the balls. I think it is only about the end result. Or you could also consider cases where you put 2 balls at the same time, cases where you take out a ball you put in earlier, etc.
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