Seven children, A, B, C, D, E, F, and G, are going to sit in seven chairs in a row. Child A has sit next to both B & G

[BTGmoderatorLU]

Source: Magoosh

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

The OA is A

[swerve]

Source: Magoosh

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

The OA is A

The first 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, the total number of arrangements is \(2\cdot 4! \cdot 5 =240\)

Therefore, A

[Scott@TargetTestPrep]

Source: Magoosh

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

The OA is A

Solution:

Since A has to sit next to both B and G, the arrangement for these three children must be either B-A-G or G-A-B.

For the first case, considering [B-A-G] as a single entity, we are actually arranging five items which are [B-A-G], C, D, E and F. There are 5! = 120 ways to arrange these five items.

The second case is similar to the first case and there are also 120 ways to arrange [G-A-B], C, D, E and F.

In total, there are 120 + 120 = 240 ways to arrange the children so that A sits next to both B and G.

Answer: A