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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, 04:36
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21
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A
B
C
D
E

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Question Stats:

69% (02:06) correct 31% (02:25) wrong based on 239 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. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800

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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, 14:43
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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. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct.

First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order.

Now, put the other five in any order — that’s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C.

Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B.

Answer = B
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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 Sep 2018, 06:05
1
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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. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800


Since A and B must occupy adjacent positions, consider AB a single element in the arrangement.
The number of ways to arrange the 6 elements AB, C, D, E, F and G = 6! = 720.

In 1/2 of these arrangements, C will be to the LEFT of AB.
In the remaining 1/2 of these arrangements,C will be to the RIGHT of AB.
Thus, the number of arrangements in which C is to the right of AB = (1/2)(720).

Since AB can switch to BA -- doubling the total number of possible arrangements -- we multiply by 2:
(2)(1/2)(720) = 720.


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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, 04:38
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


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seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194096.html
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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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 21 Oct 2015, 04:04
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A, B, C, D, E, F, G- seven children, of which A & B must sit next to each other. Considering them as one X (A, B), we have X, C, D, E, F, G. These can be arranged in 6! ways. But A,B can arrange themselves in 2! ways. So a total of 6!*2! ways = 1440. Since in exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B, therefore 1440/2 = 720.

B is the right answer.
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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 22 Oct 2015, 04:11
it is

(6!*2)/2=6!=720

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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 Jul 2017, 05:58
2
the answer must be B
say AB*5*4*3*2*1=120
4*AB*4*3*2*1=96
4*3*AB*3*2*1=72
4*3*2*AB*2*1=48
4*3*2*1*AB*1=24
total = 360 ways now as we also know that AB can be rearranged in every case in 2 ways so 360*2=720 ways

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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 21 Aug 2017, 23:00
Bunuel wrote:
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

This problem is a notch harder than anything you are likely to see on the GMAT. If you can master the principles in this problem, you certainly will be able to handle almost any problem the GMAT could concoct.

First, consider the restriction of A & B. As with problem #1 above, there are 12 possibilities for A & B, counting both position and order.

Now, put the other five in any order — that’s 120 possibilities, for a total number of configurations of 1440. That number does not take into account the restriction with C.

Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B. Therefore, in (1/2)*1440 = 720 configurations, C will be to the right of A & B.

Answer = B



Hi, I am not quite sure i get how you come to the following conclusion "Think about those 1440 configurations. In exactly half of them, C will be to the right of A & B, and exactly half, C will be to the left of A & B"
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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 22 Aug 2017, 01:40
Hey Bunuel,
can you please check this approach.

Let 7 seats be 1 2 3 4 5 6 7

Condition : Children A & B must sit next to each other, and child C must be somewhere to the right of A & B.
So let this be the right side of A & B ----> 1 2 3 4 5 6 7.

Now considering A & B as a single unit, we have 6 seats for them. (Omitting the number 1 seat as if A & B starts sitting from seat number 1, C can't be seated to their right). So we have 6C2 X 2! = 30.

Since 3 out of 7 seats are filled as C, A, B needs to be seated in this particular order always. Out of remaining 4 seats 4 members can be arranged as 4! = 24.

Hence answer would be 24 X 30 = 720.
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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 Sep 2018, 04:43
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


With A and B together as one entity total arrangements of 6 entities = 6!*2! = 720*2

in half of the case C will sit to the right of A and B and in other half cases C will sit to the left of A and B

Hence desired outcomes = (1/2)*6!*2! = 720

Answer: option B
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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 10 Sep 2018, 01:49
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


I solved it this way:

Consider 7 seats. x x x x x x x x
Place A and B together.
case 1: ABxxxxx
Here C can be arranged in 5 ways. The remaining children in 4! ways. A and B in 2! ways. So total: 2!.5.4!
case 2: xABxxxx
Here C can be arranged in 4 ways. The remaining children in 4! ways. A and B in 2! ways. So total: 2!.4.4!
Similarly we can have :
case 3: xxABxxx = 2!.3.4!
case 4: xxxABxx = 2!.2.4!
case 5:xxxxABx = 2!.1.4!

adding all cases we get: 2!.4!(5+4+3+2+1)
=2.24.5.3
=720 (B)
(I followed this approach as the total combinations when AB taken as a unit and dividing by 2 didn't strike my brain :D , so found this alternate approach. Hit kudos if you like.)

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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 Feb 2019, 18:28
Bunuel wrote:
Seven children — A, B, C, D, E, F, and G — are going to sit in seven chairs in a row. Children A & B must sit next to each other, and child C must be somewhere to the right of A & B. How many possible configurations are there for the children?

A. 600
B. 720
C. 1440
D. 4320
E. 4800



Kudos for a correct solution.


Since children A and B must always sit next to each other, we can consider them as a single entity [A-B]. We can express the row of children as follows:

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

We see that there are 6 slots which can be arranged in 6! = 720 ways.

We also have to account for the number of ways to arrange A-B, which is 2!, so the total number of arrangements is 1,440.

Now we must remember that by symmetry, the number of ways where child C is on the right of [A-B] is equal to the number of ways where child C is on the left of [A-B]; so there are 1440/2 = 720 ways in which C will be on the left and 720 ways in which C will be on the right.

Answer: B
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Re: Seven children — A, B, C, D, E, F, and G — are going to sit in seven c   [#permalink] 15 Feb 2019, 18:28
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