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In how many ways 9 identical objects can be put in non-empty boxes of arbitrary size?

Is solution integer partition of 9? That is 30?

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closed as off-topic by Did, Leucippus, Cesareo, Riccardo.Alestra, Lee David Chung Lin Jan 17 at 12:23

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    $\begingroup$ Are the boxes distinguishable? That is, is putting eight in the first box an one in the second different from putting one in the first and eight in the second? $\endgroup$ – saulspatz Jan 14 at 16:29
  • $\begingroup$ The boxes are not distinguishable. $\endgroup$ – Stratocarter Jan 14 at 16:33
  • $\begingroup$ Then it's partitions, as you said. (I haven't checked that the answer is $30$.) $\endgroup$ – saulspatz Jan 14 at 16:34
  • $\begingroup$ Putting eight in the first box an one in the second is the same as putting one in the first and eight in the second. $\endgroup$ – Stratocarter Jan 14 at 16:34
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We are talking about partitions.

https://en.wikipedia.org/wiki/Partition_(number_theory)

In your case, they are

9

8-1

7-2

7-1-1

6-3

6-2-1

6-1-1-1

5-4

5-3-1

5-2-2

5-2-1-1

5-1-1-1-1

4-4-1

4-3-2

4-3-1-1

4-2-2-1

4-2-1-1-1

4-1-1-1-1-1

3-3-3

3-3-2-1

3-3-1-1-1

3-2-2-2

3-2-2-1-1

3-2-1-1-1-1

3-1-1-1-1-1-1

2-2-2-2-1

2-2-2-1-1-1

2-2-1-1-1-1-1

2-1-1-1-1-1-1-1

1-1-1-1-1-1-1-1-1

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Assuming the boxes are distinguishable, we can put up to 8 bars between 9 objects, each bar separating two boxes. There are 2^8 ways to set the bars, 2 meaning a bar is either there or not. So the answer is 256.

Assuming the boxes are indistinguishable, the question becomes sum of number seperation. There are 1 one to use only 1 box, 8 ways to use 2 boxes, 7 ways to use 3 boxes (711,621,531,522,441,432,333), etc.

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  • $\begingroup$ The boxes are indistinguishable. But, there is only 4 ways to use 2 boxes: 81, 72, 63, 54. Am I wrong? $\endgroup$ – Stratocarter Jan 14 at 16:49
  • $\begingroup$ Oops, typo, 8 should be 4. See my counting with case of 3 boxes - same idea. $\endgroup$ – The Pianist Jan 14 at 17:08

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