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

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

27 Jan 2008, 14:31

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?

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Re: Combination of Dwarfs! [#permalink]

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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Re: Combination of Dwarfs! [#permalink]

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?

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Re: Combination of Dwarfs! [#permalink]

27 Jan 2008, 14:55

jimmyjamesdonkey wrote:

if you mean D_E and E_D, it is wrong. D_E E_D D_E does not work.
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Re: Combination of Dwarfs! [#permalink]

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?

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Re: Combination of Dwarfs! [#permalink]

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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Re: Combination of Dwarfs! [#permalink]

27 Jan 2008, 15:14

Thanks so I was on the right track! Just a little fine tuning!

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

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

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

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