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by sukhman » Thu Sep 12, 2013 8:36 am
Three dwarves and three elves sit down in a 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? Answer [spoiler]6 × 2 × 1 × 3 × 2 × 1 = 72[/spoiler]
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by sanjoy18 » Thu Sep 12, 2013 8:45 am
there would be two cases 1) dedede 2) ededed
case 1 & 2 can be done in total 36 ways each
3*3*2*2*1*1=36


hence 36+36 =72
sukhman wrote:Three dwarves and three elves sit down in a 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? Answer [spoiler]6 × 2 × 1 × 3 × 2 × 1 = 72[/spoiler]
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by vinay1983 » Thu Sep 12, 2013 9:26 am
The dwarfs and elves have to arranged in alternate manner.

Dwarfs first 3 3 2 2 1 1 so this is "and" arrangement, Hence 36

OR

Elves first 3 3 2 2 1 1 so 36 again

Total 36 or 36 i.e 36+36=72
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by [email protected] » Thu Sep 12, 2013 12:04 pm
Hi sanjoy18 and vinay1983,

You're both absolutely correct. You can save a bit of time though (and avoid the extra calculation), if you think of it this way:

With 3 dwarves and 3 elves, the seating must alternate. With the same number of each, the first "spot" could be either a dwarf or an elf.

So 6 - - - - -

The rest of the pattern proceeds normally...

6 3 2 2 1 1

(6)(3)(2)(2)(1)(1) = 72

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