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If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible?

This question seems simple using a stars and bars approach. It would seem that you should be able to partition 11 items, and place 3 partitions (or 8 boards) in $11 \choose 3$ different ways.

However, I'm seriously stumped here. I've seen this before, but no one can explain to me why the following approach is wrong:

1) Lay out your 8 boards

2) You now have 9 places to place dividers (7 in between boards, and 2 on each side)

3) The same holds true for the next 2 boards. In all, you should have (to my, incorrect, mind) $9^3$ ways of partitioning these boards among the 3 schools.

When reading off which boards ended up with each school, you would simply read left to right. All of the boards before the first divider go to school 1. Between dividers 1 and 2 would go to school 2 and so forth.

I really don't see why this is wrong, someone please convince me of my folly.

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    Suppose the first two dividers are placed between the second and third board, and between the fifth and sixth board. Or maybe it's between the fifth and sixth board, and between the second and third board. I forget. :-P How many times do you count this? Once? Or twice? – 2017-02-07
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    I follow everything except where you came up with $9^3$. The dividers are also indistinguishable, so there is no sense in trying to assign a "triple" of divider locations. They are just a $3$-subset of all the places (boards and dividers, eleven places in all). – 2017-02-07
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    @BrianTung I think I understand what you mean – 2017-02-07
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    @hardmath it's because it's 9 options 3 times – 2017-02-07
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    That's where I disagree. You cannot make a choice from nine places three times. Read my explanation, but switching the places of two dividers does not make sense. – 2017-02-07
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    @hardmath, mhm. Switching two dividers changes nothing, that's true. – 2017-02-07

2 Answers 2

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If you place the first divider between boards 1 and 2, and the second divider between boards 3 and 4, then this is really the same division of boards as when you place the first divider between boards 3 and 4, and the second divider between boards 1 and 2. In other words, your method is counting the same divisions multiple times!

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Others have correctly pointed out that changing the order of where you place the dividers doesn't matter. The naive thing would then be to just find the number of ways to select three positions of the nine, getting $9 \choose 3$, but it is allowed for a school to get no boards so you have to be able to place multiple dividers in one space. The classic technique to solve this is to add one hypothetical board to each school's take. Now you are dividing $12$ boards, which gives $11$ places to put the dividers. You cannot put dividers at the ends any more because the first and fourth schools must each get a board. You also can't put two dividers in any one slot because a middle school wouldn't get a board. That gives the $11 \choose 3$ possibilities. Now take a board away from each school and you are done.