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# Three dwarfs and three eleves sit down in a row of sit

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Director
Joined: 01 May 2007
Posts: 795

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Three dwarfs and three eleves sit down in a row of sit [#permalink]

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27 Jan 2008, 14:31
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Three dwarfs and three eleves sit down in a row of sit 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 dwarfs sit?

Kudos [?]: 379 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4583 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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27 Jan 2008, 14:39
1. $$N=C^2_1*P^3_3*P^3_3=2*3*2*3*2=72$$ if elves and dwarfs are distinguishable.

or

2. $$N=C^2_1=2$$ if elves and dwarfs are not distinguishable.
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Kudos [?]: 4583 [0], given: 360

Director
Joined: 01 May 2007
Posts: 795

Kudos [?]: 379 [0], given: 0

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27 Jan 2008, 14:41
72 is correct. Can you please tell me where I went wrong initally?

I said
6_5_4_3_2_1
D_E_D_E_D_E

Each paid of D_E is 2!, and we have three pairs. 2!*2!*2! = 8
We have three pairs 3*2*1 = 6

6*8 = 48.

Any advice where I screwed up?

Kudos [?]: 379 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4583 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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27 Jan 2008, 14:55
jimmyjamesdonkey wrote:
72 is correct. Can you please tell me where I went wrong initally?

I said
6_5_4_3_2_1
D_E_D_E_D_E

Each paid of D_E is 2!, and we have three pairs. 2!*2!*2! = 8
We have three pairs 3*2*1 = 6

6*8 = 48.

Any advice where I screwed up?

if you mean D_E and E_D, it is wrong. D_E E_D D_E does not work.
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Kudos [?]: 4583 [0], given: 360

Director
Joined: 01 May 2007
Posts: 795

Kudos [?]: 379 [0], given: 0

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27 Jan 2008, 15:07
Good catch...so I'm redoing..

6_5_4_3_2_1
D_E_D_E_D_E

I can arrange the D's 3! and the E's 3! to give me 3!*3! = 36. How do I get to 72 from here?

Kudos [?]: 379 [0], given: 0

CEO
Joined: 17 Nov 2007
Posts: 3584

Kudos [?]: 4583 [0], given: 360

Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40

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27 Jan 2008, 15:10
jimmyjamesdonkey wrote:
Good catch...so I'm redoing..

6_5_4_3_2_1
D_E_D_E_D_E

I can arrange the D's 3! and the E's 3! to give me 3!*3! = 36. How do I get to 72 from here?

It is simple:

D_E_D_E_D_E
E_D_E_D_E_D
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Kudos [?]: 4583 [0], given: 360

Director
Joined: 01 May 2007
Posts: 795

Kudos [?]: 379 [0], given: 0

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27 Jan 2008, 15:14
Thanks so I was on the right track! Just a little fine tuning!

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Re: Combination of Dwarfs!   [#permalink] 27 Jan 2008, 15:14
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