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If I have 20 disks of one color, let's say black. And 20 of another, let's say white. And I want to make stacks. The amount of disks in the stack can range from 1 to 40. But the colors cannot be mixed in the sense that all black disks appear consecutively and all white disks appear consecutively. Any number of black or white can be at the bottom or top. But never mixed. How many possible stacks can I make?

JimN
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Jeroen
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  • Imagine you have five disks, two white and three black. Normally you might expect ten possibilities: $WWWBB$, $WWBWB$, $WBWWB$, $BWWWB$, $WWBBW$, $WBWBW$, $BWWBW$, $WBBWW$, $BWBWW$ and $BBWWW$, where the two extremes need to be removed (WWWBB and BBWWW), so are left with eight possibilities. Is that correct? – Dominique Dec 20 '23 at 10:18

1 Answers1

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Let's say you can only have white disks stacked on the bottom, that means there are 20 possible white stacks on the base of the tower.

For each of these possibilities we have 21 possible stacks of black disks on top. That is up to 20 black disks plus the possibility of having no black disks.

So we have 20 times 21 which is 420 possible stacks. Multiplying this by 2 to account for the case where black discs are placed on the bottom, we have a total of 840 possible stacks.

rangen1
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  • Thanks. Sounds right to me. – Jeroen Dec 20 '23 at 15:01
  • Hmmm. But now I'm thinking wouldn't the extra 1 of having no black disks be one of the 20 white options? – Jeroen Dec 20 '23 at 15:17
  • @Jeroen Another way to phrase this might be to consider first the stack of one white disk, count the possible stacks that have only one white disk at the start, that is: 1 possibility with no black disks, 1 whith 1 black disk,1 with 2 and so on, this gives us 21 possibilities altogether. Now repeat this with the other 19 possible white stacks at the bottom and add all these possibilities together (21, 20 times) to get 420 as desired. Hope this answers your question. – rangen1 Dec 23 '23 at 14:45