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

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Senior SC Moderator
Joined: 14 Nov 2016
Posts: 1348
Location: Malaysia
Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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22 May 2017, 07:59
9
00:00

Difficulty:

35% (medium)

Question Stats:

66% (01:17) correct 34% (01:15) wrong based on 179 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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22 May 2017, 08:13
1
3
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

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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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22 May 2017, 09:12
1
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

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Joined: 23 Jul 2014
Posts: 81
Re: Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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29 Aug 2018, 06:04
2

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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Joined: 03 Dec 2018
Posts: 9
Re: Three gnomes and three elves sit down in a row of six chairs. If no gn  [#permalink]

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14 Mar 2019, 02:46
1
anud33p wrote:

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!

Hi,
I hope it is not too late.
The error in this is approach is that by performing 4C3 you are selecting any 3 spaces out of 4 which is not be the desired requirement.

Let's say that we have already arranged the Gnomes.

_G_G_G_
1 2 3 4

Now we have 4 places where the elves can sit. By performing 4C3 we are selecting any 3 spaces.

But what if we select 1st,3rd and 4th number spaces for the Elves to sit?

That is :

E G G E G E
1 3 4

This is wrong as the question requires the Gnomes not to sit along each other.

Hence the best approach for this question would be :
1. EGEGEG: 3!x 3! =36
2. GEGEGE: 3!x 3! =36

total ways=36+36=72

Hope that helps.
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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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14 Jun 2019, 15:46
1
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

We can see that one seating arrangement of gnomes (G) and elves (E) can be:

GEGEGE

The first G and E each has 3 choices; the second G and E each has 2 choices and the last G and E each has 1 choice. Thus the number of ways to have the “GEGEGE” seating arrangement is:

3 x 3 x 2 x 2 x 1 x 1 = 36

However, the seating arrangement can also be EGEGEG, and there will also be 36 such arrangements. Thus, the total number of seating arrangements is 36 + 36 = 72.

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Re: Three gnomes and three elves sit down in a row of six chairs. If no gn   [#permalink] 14 Jun 2019, 15:46
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