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3 dwarves and 3 Elves sit down in a row of 6 chairs. If no

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3 dwarves and 3 Elves sit down in a row of 6 chairs. If no [#permalink]

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New post 28 Dec 2007, 10:41
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3 dwarves and 3 Elves sit down in a row of 6 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?

I feel the approach in MGMAT for this problem is not the best ... :roll: So I am looking for alternatives...Thanks..
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Re: Combinatorics - Dwarf and Elves [#permalink]

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New post 28 Dec 2007, 10:46
Beyond700 wrote:
3 dwarves and 3 Elves sit down in a row of 6 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?

I feel the approach in MGMAT for this problem is not the best ... :roll: So I am looking for alternatives...Thanks..


S S S S S S
D E D E D E Possible seat for D 3C1=3 Possible seat for E 3C1=3 9 ways
E D E D E D same result 9

Total 18 ways
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New post 28 Dec 2007, 10:51
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dedede: N1=3P3*3P3=6*6=36
ededed: N2=3P3*3P3=6*6=36

3P3 - 3 different things at 3 different positions.

N=36*2=72
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New post 28 Dec 2007, 11:09
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walker wrote:
72

dedede: N1=3P3*3P3=6*6=36
ededed: N2=3P3*3P3=6*6=36

3P3 - 3 different things at 3 different positions.

N=36*2=72


Bulls eye....

But I did this in this way (simple layman terms)

'Chairs ' - 1 2 3 4 5 6
possibile - 6*3*2*2*1*1 = 72

The good thing is that I managed to solve 4 to 5 such questions and but I am not sure how efficient this approach will be.. Any comments...

MGMAT has 1/2 page solution for this problem...
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New post 28 Dec 2007, 11:44
Beyond700 wrote:
The good thing is that I managed to solve 4 to 5 such questions and but I am not sure how efficient this approach will be.. Any comments...


maybe this will be useful: http://www.gmatclub.com/forum/t56486

I think it is not a good idea to use only one approach for combination-permutation-probability problems.
I have a few general principles that seems be helpful for me in CPP problems.

1. try to find a answer by several ways.
2. use pattern approach for enumeration of possibilities for problems with complex restrictions.
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Re: Combinatorics - Dwarf and Elves [#permalink]

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New post 27 Sep 2009, 10:26
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3 dwarves and 3 Elves sit down in a row of 6 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?

Soln:
Assuming the Dwarves taken 1st , 3rd and 5th place. The other 3 places will be taken by Elves.
Hence total number of arrangements = 3! * 3!

Now if Dwarves take 2nd, 4th and 6th place. The other 3 places will be taken by Elves.
Hence total number of arrangements = 3! * 3!

Thus total number of ways is = 3! * 3! + 3! * 3! = 72 ways
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Re: 3 dwarves and 3 Elves sit down in a row of 6 chairs. If no [#permalink]

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Re: 3 dwarves and 3 Elves sit down in a row of 6 chairs. If no   [#permalink] 07 Nov 2016, 16:56
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