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

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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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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Similar yet different questions:
six-children-a-b-c-d-e-and-f-are-going-to-sit-in-six-chairs-i-194098.html
seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194097.html
seven-children-a-b-c-d-e-f-and-g-are-going-to-sit-in-seven-c-194096.html
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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.

it is

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

B

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

if you liked it please press kudos

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"

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.

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

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.)

Thanks,
Uma

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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Here is it...

Case 1: Since AB must be together, we have AB _ _ _ _ _ = 6! We can also have BA _ _ _ _ _ = 6!
In the absence of any restriction, the maximum arrangements we can have is 1,440. ( Option C, D & E are out)
Because of the restriction, we'll proceed to case 2.

Case 2: Using the slot method: we can have; AB _ _ _ _ C, _ AB _ _ _C , _ _ AB _ _ _ C , _ _ _ AB_ C, _ _ _ _AB C.
C all to the right of AB. So also we can have it all to the left of AB.
Therefore, C stands a 50% possibility of being arranged to the right of AB and 50% possibility of being arranged to the left of AB.
Hence, 50% (arrangement to the right of AB) of 1,440 = 720.