Three dwarves and three elves sit down in in the row of six

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Three dwarves and three elves sit down in in the row of six [#permalink]

19 Jan 2006, 13:06

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Three dwarves and three elves sit down in in the row of six chairs. If no dwarf will sit next to another dwarf and no elf will sit next to another elf, in how many different ways can the elves and dwarves sit.
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19 Jan 2006, 15:13

two posibilities

DEDEDE
or
EDEDED

so 3!*3! + 3!*3! = 72

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19 Jan 2006, 18:40

The elves and dwarves must alternate:

e d e d e d or d e d e d e

Considering the first case, the total number ways is to rotate the elves 3! times while keeping the dwarves to one combination. Then rotate the dwarves 3! times while keeping the elves to one combination.

However, that is only for the first case. We must multiply by 2 as we need to account for the second case.

Total = 3! * 3! * 2 = 72

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25 Jan 2006, 17:13

3! * 3 ! for DEDEDE
3! * 3 ! for EDEDED

Total = 36+36 = 72.

[#permalink] 25 Jan 2006, 17:13

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Three dwarves and three elves sit down in in the row of six

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Three dwarves and three elves sit down in in the row of six

Three dwarves and three elves sit down in in the row of six

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