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Seven children — A, B, C, D, E, F, and G — are going to sit in seven c

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Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 04 Mar 2015, 04:34
1
7
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A
B
C
D
E

Difficulty:

  35% (medium)

Question Stats:

68% (01:32) correct 32% (02:03) wrong based on 290 sessions

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Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 04 Mar 2015, 04:40
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


Kudos for a correct solution.



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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 04 Mar 2015, 09:50
1
In order to sit together they become one entity that instead of having 7 students we have 4+ group of 3 = 4+1= 5 entities. In order to arrange 5 = 5! and the group of 3 can be arranged in 2 ways GAB or BAG hence total possible configurations 240 answer is A.
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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 08 Mar 2015, 13:43
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

First, we will consider the restricted elements — children A & B & G have to be in three seats in a row. How many “three in a row” seats are there in a row of seven seats?

X X X _ _ _ _

_ X X X _ _ _

_ _ X X X _ _

_ _ _ X X X _

_ _ _ _ X X X

There are five different “three in a row” locations for these three children. Now, for any given triplet of seats, we know A has to be in the middle, so the children could be seated B-A-G or G-A-B — just those two orders. This means the total number of configurations for these three children is 5*2 = 10.

Now, consider the non-restricted elements, the other four. Once A & B & G are seated, the remaining four children can be seated in any order among the remaining four seats — that’s a permutation of the 4 items —- 4P4 = 4! = 24. For any single configuration of A & B & G, there are 24 ways that the other children could be seated in the remaining seats.

Finally, we’ll combine with the Fundamental Counting Principle. We have 10 ways for the first three, and 24 ways for the remaining four. That’s a total number of configurations of 24*10 = 240.

Answer = A
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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 18 Mar 2016, 01:21
1 case is: BAGCDEF.
Now B and G can be arranged in 2 ways among themselves. C,D,E and F can be arranged in 4! Ways. A can take 5 positions.
So total number of arrangements is 2*4!*5 =240 . Hence A.
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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 15 Jun 2017, 05:36
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


Kudos for a correct solution.


Since A has to sit in between B & G so arrangement of these 3 can be BAG & GAB.
Also apart from these 3, other 4 children can sit in any order.. SO possible configuration of arrangement = 2 x 5! = 2 x 5 x4 x3 x2 x1 = 240

Answer A.

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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 06 Jul 2017, 17:25
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


We can express the row of children as follows:

[B-A-G][C][D][E][F]

We see that there are 5 slots which can be arranged in 5! = 120 ways.

We also have to account for the number of ways to arrange B-A-G such that A is always in the middle of B and G, and there are two ways to make that arrangement (B-A-G and G-A-B).

Thus, there are 2 x 120 = 240 possible configurations.

Answer: A
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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c  [#permalink]

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New post 03 Nov 2017, 09:01
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Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Child A has to sit next to both B & G, with these two children immediately adjacent to here on either side. The other four children can sit in any order in any of the remaining seats. How many possible configurations are there for the children?

A. 240
B. 480
C. 720
D. 1440
E. 3600


Take the task of seating the 7 children and break it into stages.

We’ll begin with the most restrictive stage.

Stage 1: Arrange children A, B and G
Since child A has to sit next to both B & G, we can conclude that child A must sit BETWEEN B and G
There are only 2 options: BAG and GAB
So, we can complete stage 1 in 2 ways

IMPORTANT: Once we've arranged A, B and G, we can "glue" them together to form a single entity. This will ensure that A is between B and G

Stage 2: Arrange the single entity and the four remaining children
There are 5 objects to arrange: C, D, E, F and the BAG/GAB entity.
We can arrange n different objects in n! ways
So, we can arrange the 5 objects in 5! ways (5! = 120)
So, we can complete stage 2 in 120 ways

By the Fundamental Counting Principle (FCP), we can complete the 2 stages (and thus arrange all 7 children) in (2)(120) ways (= 240 ways)

Answer:

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c   [#permalink] 03 Nov 2017, 09:01
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