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

[M7MBA]

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 her 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

Answer: A

Source: Magoosh

[Brent@GMATPrepNow]

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 her 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

Answer: A

Source: Magoosh

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: A