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Three gnomes and three elves sit down in a row of six chairs. If no gn

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Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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New post 22 May 2017, 07:59
4
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (01:21) correct 38% (01:18) wrong based on 95 sessions

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Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

A. 18
B. 36
C. 48
D. 72
E. 96

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Re: Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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New post 22 May 2017, 08:13
hazelnut wrote:
Three gnomes and three elves sit down in a row of six chairs. If no gnome will sit next to another gnome and no elf will sit next to another elf, in how many different ways can the elves and gnomes sit?

A. 18
B. 36
C. 48
D. 72
E. 96


Case 1: G E G E G E - Total ways = 3!*3! = 36

Case 2: E G E G E G - Total ways = 3!*3! = 36

Total Ways of arrangements = 36+36 = 72

Answer: Option D
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Re: Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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New post 22 May 2017, 09:12
Basically the above objective would be achieved if elves and genomes sit alternately.
Let the chair numbers be 1, 2, 3, ...6.
So we can make them sit in two ways:

Case 1. Elves on chair numbers 1, 3, 5. Genomes on chair numbers 2, 4, 6
This can be done in 3! * 3! = 36 ways

Case 2
. Genomes on chair numbers 1, 3, 5. Elves on chair numbers 2, 4, 6
This can be done in 3! * 3! = 36 ways

Thus total ways = 36+36 = 72

Hence answer is D
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Re: Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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New post 29 Aug 2018, 06:04
chetan2u Bunuel mikemcgarry : Could you please help me find the glitch in my approach.

I understand the logic mentioned above but I would like to know what is wrong with:

1. First placing three Gnomes: G G G
2. Now, there are four places to fit in three Elves. Number of ways to select three out of four places = 4C3
3. Permute the three genomes and three elves = 4C3 * 3! * 3!

By this approach, the answer I get is 144. What is wrong? o. O

Thank you!
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Re: Three gnomes and three elves sit down in a row of six chairs. If no gn &nbs [#permalink] 29 Aug 2018, 06:04
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