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There are $3$ elves and $3$ dwarves and $6$ chairs. The elves and dwarves are trying to integrate with each other and will only sit next to someone of the opposite race and not next to their own kind. How many arrangements are there for them to sit together.

So I think I have $1$ solution. Is there a more elegant way to do this?

So there are $6$ chairs. The first person has a choice of $6$, but the last $2$ of his type are restricted in their seating; they can only have $2$ choices, and then 1 choice after that. For example, if an elf sits in seat 1, the second elf can only either sit in seat $3$ or seat $5$, and the last elf can only sit in the remaining seat. Once the dwarfs sit, they only have 3 remaining seats to choose from. So I think the math is:

$6*2*1*3*2*1 = 72$.

Is there a more elegant way to do this? Since the arrangements either look like EDEDED or DEDEDE, can I start from there?

Ski Mask
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Jwan622
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  • How are either of those possible arrangements? The problems states that elves and dwarves won't sit together. EDEDED and DEDEDE have them sitting next to each other! Doesn't the problem mean something more like EEEDDD so the elves can sit with elves and dwarves with dwarves. Then just permute among the dwarves and elf group respectively? – mathematics2x2life Feb 05 '14 at 06:29
  • I'm a fool, I changed the wording of the problem. – Jwan622 Feb 05 '14 at 19:09

1 Answers1

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You can. Suppose it's EDEDED. Then there are $3!=6$ ways to arrange the elves, and $3!=6$ ways to arrange the dwarves. Multiplying these, we get $36$ arrangements that look like EDEDED. There are $36$ more for DEDEDE, and that makes $72$.

G Tony Jacobs
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