# Combinatorics distribution into urns

I have solved the following question and attached my solution. But I feel my solution is sort of lengthy.Is there a better, more concise way possible to solve this question?

There are 6 identical white balls and 6 identical black balls. They have to be distributed among 10 (different) urns such that there is at least 1 ball in each urn. How many ways can this be done?
1. 25000
2. 26250
3. 28250
4. 13125

option 2 is the correct answer. Here is a PDF of my solution. It's in handwriting so kindly excuse any legibility problems.

Since you have $12$ balls and $10$ urns, either one urn gets $3$ balls, or $2$ urns get $2$ balls each. I’d deal with the two cases separately, as you did.

• Suppose that one urn gets $3$ balls; there are $10$ ways to choose that urn. Put $3$ white balls into it and one white ball into each of the other $9$ urns. Now we have to choose $6$ of the $12$ white balls to be painted black. Since balls of the same color in the same urn are indistinguishable, all we really need to know is which of the singletons are black. If $4$ singletons are black, for instance, the urn with $3$ balls contains $2$ black balls and one white ball. Thus, it’s really a matter of choosing $3,4,5$, or $6$ of the singletons to paint black; if we chose fewer than $6$, the missing black balls are in the urn with $3$ balls. This case therefore accounts for $$10\left(\binom93+\binom94+\binom95+\binom96\right)=2\cdot10\left(\binom93+\binom94\right)=4200\;.$$

• Now suppose that $2$ urns get $2$ balls each; there are $\binom{10}2=45$ ways to choose the lucky pair of urns. Again start with $12$ white balls and choose $6$ to paint black, organizing the subcases according to the number of singletons painted black. If $2$ or $6$ are painted black, no further choices are possible: we know what has to be in the two urns containing $2$ balls each. If $3$ or $5$ are painted black, one of the $2$ urns with $2$ balls contains two of the same color, and the other contains one of each color, so there are $2$ distinguishable arrangements of the balls in the urns with $2$ balls each. And if $4$ singletons are painted black, there are $3$ distinguishable arrangements of the balls in the urns with $2$ balls each. Finally, $\binom82=\binom86$ and $\binom83=\binom85$, so the grand total for this case is $$45\left(2\binom82+4\binom83+3\binom84\right)=22050\;.$$

The final total is then $4200+22050=26250$. This is really only a slightly more efficient version of your analysis.

Basic Assumption: all the urns are distinct.

As above, break the approach in two parts. One is when there are 3 balls in a single urn and the other when there are 2 balls in 2 urns.

Case I : As there is one single urn, we can choose that in $C(10,1)$ ways. Now, that urn can have $4$ different type of combinations:

• BWW
• BBW
• BBB
• WWW

By similarity, the first two will have same number of combinations as likely as the second two.

So number of ways: $C(10,1) \times {2 \times C(9,4) + 2 \times C(9,3)}.$ First choosing $4$ urns out of remaining $9$ urns to place $4$ remaining white balls in the first option and 4 black balls in the second option. They are same so multiplied by $2$. In another, choosing $3$ urns from remaining $9$ urns so as to place $3$ remaining black balls in the third and 3 remaining white balls in the fourth.

Case II : Here,the cases can be

1. WB,WB
2. WW,BB
3. WB,BB
4. WB,WW
5. BB,BB
6. WW,WW

Applying the above approach :

$$C(10,2){4\times C(8,3) + 3\times C(8,4) + 2\times C(8,2)}.$$

Tell me if you didn't understand how we got to the last equation.