# Distributing colored balls in 80 cells so that no two balls of the same color are in the same cell

An urn contains 50 white balls and 50 black balls(balls are indistinguishable). We want to distribute all the balls into 80 numbered cells, where at-most 1 ball of the same color is allowed to occupy any cell. Find the probability that all the cells are occupied.

We all got different answers for this problem. Some answer was only a simple fraction using binomials, some had big summations involving product of three binomials in the denominator.

I would be grateful if someone outlines the solution and if possible provides the correct answer.

My attempt

At first I counted the number of ways to place the balls so that all the cells are occupied. By PHP, we know that a cell can not contain $$\ge 3$$ balls. Let there be $$a$$ cells containing 2 balls. Then we know that $$100=2a+(80-a)\implies a=20$$. Now we choose which 20 cells contains 2 balls. This can be done in $$\binom{80}{20}$$ ways. Now we have to distribute the remaining 60 balls in 60 cells. But these remaining 60 balls consist of 30 black and 30 white balls. Thus we only need to choose which 30 cells we want to put the white balls in. So in total there are $$\binom{60}{30}\cdot \binom{80}{20}$$ ways.

Now I did this same procedure assuming that the balls will e distributed in $$k$$ cells, $$50\le k\le 80$$. This gave me really huge sum which I could not simplify and don't remember now. If I recall it, I will add it asap. But I wanted to know whether my method is at-leats correct?

– lulu
Commented Apr 1, 2022 at 13:32
• Ok, I will add my attempt. Commented Apr 1, 2022 at 13:33
• Note that, as it stands, the question is not clear to me. How does probability enter? Where, exactly, is there randomness in this system and what distribution is followed? I suspect that the different answers come about because people are modeling the randomness differently.
– lulu
Commented Apr 1, 2022 at 13:34
• Hi, could you specify which part of the problem is unclear? I will try to clarify it to the best of my ability. Commented Apr 1, 2022 at 13:45
• Are you assuming that the white and black balls are placed independently? And that each admissible arrangement of the balls of a single color are equally probable?
– lulu
Commented Apr 1, 2022 at 13:47

Using indifference here, where any of $$\binom{80}{50}$$ possible distribution are equally likely. From the rule, there will be 20 cells that contain both colours, then 30 cells contain one white ball each and the remaining 30 cells contain one black ball each.

• Choose the 20 cells, $$\binom{80}{20}$$ possibilities
• Choose 30 of the remaining 60 cells to contain white ball, $$\binom{60}{30}$$ possibilities

So the probability is $$\frac{\binom{80}{20}\times\binom{60}{30}}{\binom{80}{50}^{2}}$$

• For avoidance of doubt: it is worth remarking that this result coincides with my result, though the expressions may appear to be different.
– lulu
Commented Apr 1, 2022 at 13:55

There are $$\binom {80}{50}$$ ways to place the balls of a single color. Let us imagine that they are all equally probable.

Fix the arrangement of the white balls (they are all the equally likely, so this choice makes no difference). Now there are $$30$$ empty cells. We need those to be occupied by black balls. Thus, to specify a "good" arrangement of black balls is the same as specifying a choice of $$20$$ cells out of the $$50$$ white occupied cells. Thus the desired probability is $$\binom {50}{20}\Big / \binom {80}{50}$$

Note that, unsurprisingly, this is effectively $$0$$.