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

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

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.

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

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

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.

RELATED VIDEOS

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.