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I’ve been working on problems similar to the one in the title for a few days, and I came up with two methods that I believe should both work, but get completely different answers. Clearly, only one of them is correct, but I don’t understand why the incorrect method doesn’t work. Here are the methods:

  1. Instead of looking at ways to pick six rocks from the line of twenty, consider a line of 14 rocks to which we must add six such that no two are consecutive. We can put up to 1 rock in each spot at either end of the line of 14, or in between each pair of rocks, for a total of 15 spots. We then choose 6 of those 15 spots to put our rocks. There are 15choose6 ways to make that choice.

  2. begin with the six rocks we pick. Fix 5 additional rocks such that none of the six rocks we pick are next to each other (since rocks are identical it does not matter which 5 we choose, but we must have these rocks in any final arrangement). We then find how many ways we can add the remaining 9 rocks to the array. For each rock, we can put the rock at either end of the array, or between two of the rocks we pick (since they’re identical, it doesn’t matter if we add to the left or right of the 5 fixed non-picked rocks) for a total of 7 choices. Each of the 9 remaining rocks can independently be placed in any of those 7 spots, providing 9*7 possible possible placements for the 9 rocks.

Clearly, 15choose6 is not the same as 63, but why do these methods count different things? That is, what possible arrangement can be achieved by the first method but not the second, or what arrangements does the first method over-count?

basjer
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  • I have no idea what you mean by "Fix five additional rocks ..." – Thomas Andrews Sep 21 '24 at 22:12
  • As a general tip for this sort of thing: Look at a smaller case where you can list everything out (say picking 3 rocks from 6, no two consecutive). Look at what the analogue of each of your two approaches says for this case, and see if you can find the reason the wrong one doesn't match your list (usually it'll either be something in your list that isn't counted at all, or two things you've been treating as different that are actually the same thing on your list) – Kevin P. Costello Sep 21 '24 at 22:31
  • @ThomasAndrews By “fix” I mean fix the position. We know that any valid arrangement of rocks that we construct must have at least one rock in between each rock we pick, so the second method says “we know we have to pick six rocks, and there has to be a rock between each of those six rocks, so every possible arrangement must share that characteristic. So let’s start with the 11 rocks we must have, then find all the ways we can freely add the remaining nine rocks to fill out the array” – basjer Sep 22 '24 at 04:11
  • @KevinP.Costello Thanks for this. I was too wrapped up in the intuition and methodology that I forgot to work it out on a smaller case. The 7*9 should be 7^9, as each placement has its own tree of successive placements, among which it is very easy to find the overcounting. This helped me understand why counting partitions is so tricky. I was completely misinterpreting my own counting. – basjer Sep 22 '24 at 04:37
  • "Each of the 9 remaining rocks can independently be placed in any of those 7 spots" and from this you decide that there are $9\times7$ total ways to place those rocks? What is the reasoning behind that? – David K Sep 23 '24 at 05:23

2 Answers2

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Well, the first method is well known, if you want an alternate procedure where you first place the $6$ "special" rocks together (red, say, against $14$ white rocks ), and then work out the answer, here's a method using "successive placement":

  • Place the $6$ red rocks together, and use up $5$ white rocks to separate them, and then just forget about these five separators. $9$ white rocks remain that can be placed one by one in the interstices which will increment by one with each placement, thus $7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15$ ways, but as we can't distinguish between the $9$ divide by $9!$, giving the answer

$$\dfrac{7\cdot8\cdot9\cdot10\cdot11\cdot12\cdot13\cdot14\cdot15}{9!} \equiv \binom {15}{6}$$

PS:

Although the "normal" method is simpler here, where various constraints are there, "successive placement" can be very useful, as ,for example, here, or much simpler, as here

true blue anil
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Each of the 9 remaining rocks can independently be placed in any of those 7 spots ...

So you have nine identical rocks to distribute into seven distinct bins. If $r_k$ is the number of rocks in the $k$th bin, you must have

$$ r_1 + r_2 + r_3 + r_4 + r_5 + r_6 + r_7 = 9. $$

That is, you're looking for the number of ways to add seven non-negative integers together and get the sum 9. This has a well-known solution by a method called "sticks and stones" (originally popularized as "stars and bars"), and the answer in this case is

$$\binom{9+7-1}{7-1} = \binom{15}{6} . $$

So there is nothing wrong with your model of counting the arrangements all the rocks by counting arrangements of just these nine rocks. The error is in how you counted the arrangements of the nine rocks.

David K
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