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The question states that

6 white and 6 black balls of the same size are distributed among 10 different urns. Balls are alike except for the colour each urn can hold any number of balls. Find the number of different distribution of the balls so that there is atleast 1 ball in each urn.

Now the difficulty I am finding is that I can't simply apply any method for a direct answer as there will be many cases which could be repeated and some which I have to add again so I thought it won't be a good method

So I am seeking a quick answer to the question with some explanation as this question is something I can't understand much..

N. F. Taussig
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Guess
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  • Since $12$ is only slightly greater than $10$, you can quickly enumerate the possible patterns (there aren't many). Then you can count the ways to populate each. – lulu Mar 26 '24 at 11:40
  • I can probably do that but what for a general case like n white balls and n Black into n different urns – Guess Mar 26 '24 at 11:44
  • For large numbers, there'd be no way to avoid a messy, Inclusion-Exclusion sort of, sum over cases. Even ignoring the color issue, the number of surjections is fairly messy, see this question. The color issue makes it worse. – lulu Mar 26 '24 at 11:46
  • To Guess : with respect to your partially more general question, where there are two distinct colors, see this answer. The answer's addendum discusses the more general $~K$-color problem. – user2661923 Mar 27 '24 at 03:02
  • @lulu For what it's worth, with respect to the partially more general question, with two distinct colors, as analysed here, I found inclusion-exclusion to be messy. So, I used the direct approach instead. See also the answer's addendum, which discusses the difficulties in the more general $~K$-color problem. – user2661923 Mar 27 '24 at 03:06

2 Answers2

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First, we divide all patterns of how balls can be distributed into two groups:

  • group of patterns A, when an urn has $3$ balls in it,

  • and group if patterns B, when two urns have two balls in it.

Examine the group A, there are following patterns in it:

  • $3$ white balls in a “big” urn,

  • $2$ white, $1$ black balls,

  • $1$ white, $2$ black,

  • $3$ black.

What is left for the other $9$ urns in these cases?

  • $3$ white, $6$ black balls,

  • $4$ white, $5$ black,

  • $5$ white, $4$ black,

  • $6$ white, $3$ black.

We can calculate the number of ways how to distribute these $9$ balls between $9$ urns. Each urn has exactly one ball, so the answer is $9\choose3$ in the first and fourth case and $9\choose4$ in the second and third.

There are $10$ ways to choose a big urn. So the number of ways in the first group of patterns is $$10\times\left(2\times{9\choose3}+2\times{9\choose4}\right)=$$

$$=10\times(168+252)=4200.$$

Now, let us consider the group B. We choose two big urns ($10\choose2$ ways to do that). Then there are the following patterns of what can be in these two big urns:

  • $4$ black balls,

  • $3$ black balls and $1$ white ball ($2$ ways to choose an urn for it),

  • $2$ black and $2$ white balls ($3$ ways to choose how they are distributed between two big urns; black balls can go $2-0$, $1-1$, or $0-2$),

  • $1$ black, $3$ white ($2$ ways),

  • $4$ white balls.

The other $8$ urns have the following $8$ balls in these cases:

  • $2$ black, $6$ white,

  • $3$ black, $5$ white,

  • $4$ black, $4$ white,

  • $5$ black, $3$ white,

  • $6$ black, $2$ white.

The number of ways to distribute these $8$ balls are $\binom82$, $\binom83$, $\binom84$, $\binom83$, $\binom82$, respectively.

Then the number of ways is

$$\binom{10}2\times\left( 2\times\binom82+ 2\times2\times\binom83 +3\times\binom84\right)=$$

$$=45\times(56+224+210)=45\times490=22050.$$

The final answer is then

$$ 4200+ 22050=26250.$$

Aig
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Here is a solution using inclusion/exclusion.

First, if we want to place $k$ balls in $r$ urns, without any restrictions, this can be done in $$\binom{k+r-1}{r}$$ ways. This fact can be shown by the method of Stars and Bars, or see the Wikipedia article on Multisets.

Now consider placing $6$ white balls and $6$ black balls in $10$ urns. Let's say an arrangement has "Property $i$" if urn $i$ is empty, for $1 \le i \le 10$, and define $S_j$ to be the total of the number of arrangements with $j$ of the properties, for $0 \le j \le 9$. Then $$S_j = \binom{10}{j} \binom{10-j+6-1}{6}^2$$ By the Principle of Inclusion/Exclusion, the number of arrangements with none of the properties, i.e. with no empty urn, is $$\sum_{j=0}^9 (-1)^j S_j = 26250$$

awkward
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