BTGModeratorVI wrote: ↑Sun Jul 26, 2020 6:40 am
Six children, Arya, Betsy, Chen, Daniel, Emily, and Franco, are to be seated in a single row of six chairs. If Betsy cannot sit next to Emily, how many different arrangements of the six children are possible?
A. 240
B. 480
C. 540
D. 720
E. 840
Answer:
B
Source: Kaplan
Solution:
We are given that Arya, Betsy, Chen, Daniel, Emily, and Franco are to be seated in a single row of six chairs, and that Betsy cannot sit next to Emily. We need to determine how many different arrangements of the six children are possible.
We can use the following equation:
[Total number of arrangements] = [number of arrangements with Betsy next to Emily] + [number of arrangements with Betsy not next to Emily].
Let’s rearrange the above equation as:
[Number of arrangements with Betsy not next to Emily] = [total number of arrangements] - [number of arrangements with Betsy next to Emily].
Since we are arranging 6 children, they can be arranged in 6! = 720 ways. Now we can determine how many ways to arrange them when Betsy is next to Emily. We can represent this arrangement as:
[A]-[B-E]-[C]-[D]-[F]
Notice that since Betsy is next to Emily, we consider them as 1 unit. Since there are 5 units to arrange, those units can be arranged in 5! = 120 ways. We also must account for the fact that when sitting together, Betsy and Emily can be arranged in 2! ways ([B-E] or [E-B]), so the total number of ways to arrange the children when Emily is next to Betsy is 2 x 120 = 240 ways.
Thus, the number of ways to arrange the children when Betsy is not next to Emily is 720 - 240 = 480 ways.
Answer: B
