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I've tried thinking about this multiple times, and it seems like I am still in some fallacy. The task is to calculate all possibilites of 35 students receiving 100 non-distinguishable balls with every student having at least one.

Here's how I went at it from a thought process:

First I split the task:

  1. 35 students receiving 35 balls with everyone having at least one
  2. 35 students receiving 65 balls with no limitation

The solution as a follow up would be 1. and 2. multiplied.

  1. is trivial. There is only 1 possibility, after all the balls are identical and there is only 1 way all 35 students can get one ball each.

  2. is a bit of a different though process. Here's the formula I'd use to calculate it:

$$\frac{35^{65}} {65!}$$

Here's my thought process: We first calculate all possible combinations assuming they are distinguishable. That would be 35 to the power of 65, since we have 35 possibilties for each ball. Afterwards we divide by all the possible "ball switching" to make the balls indistingusihable. That would be 65!.

Now, running this through Wolfram Alpha has shown me that the formula is incorrect, because we get a rational number, no integer.

I'm less interested in the correct formula to calculate it, but rather where my logical fallacy is. I hope you guys can help.

Joe
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2 Answers2

1

Consider the situation where the labelled balls are distributed with frequencies $(31,1,1,\ldots,1)$.

The number of arrangements with this distribution is $65!/31!$ since we can permute the order of the first person's balls without changing who gets which balls.

Your formula instead implies all $65!$ permutations of the ball labels gives different who-gets-which-balls arrangements. But that's not the case.

Rebecca J. Stones
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  • Could you please elaborate on the last sentence? I think I've struck a wall at that thought. – Joe Oct 21 '15 at 09:58
  • You write: "Afterwards we divide by all the possible "ball switching" to make the balls indistingusihable. That would be 65!." But it's not 65!; it's actually less than that since some ways of ball switching give the same unlabelled arrangement. (E.g. if the first person has balls 1 and 2, and you switch balls 1 and 2, then the first person still has balls 1 and 2.) – Rebecca J. Stones Oct 21 '15 at 10:01
  • I think I see what you mean. My thought process was "I need to distribute 65 balls to 65 open spots." But I think I got the correct way to think about it now, the whole thing, namely 99 choose 34, trying to seperate the 100 balls with 34 "borders". After checking my formula in the head with the real one I can confirm that it seems to be correct. – Joe Oct 21 '15 at 10:22
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Part $1$ is correct.

Part $2$ is a "star-and-bars" problem. There are $65$ "stars" (balls) and you have to create $35$ partitions (allocation to the students) by inserting $34$ "bars". Partitions can have no stars in which case two bars are next to each other. The number of combinations is $$\binom {65+34}{34}=\binom {99}{34}$$

Hypergeometricx
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