Source: Magoosh
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
A. 1/30
B. 1/15
C. 1/5
D. 2/5
E. 7/20
The OA is D
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is
This topic has expert replies
-
- Moderator
- Posts: 2207
- Joined: Sun Oct 15, 2017 1:50 pm
- Followed by:6 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Scott@TargetTestPrep
- GMAT Instructor
- Posts: 7242
- Joined: Sat Apr 25, 2015 10:56 am
- Location: Los Angeles, CA
- Thanked: 43 times
- Followed by:29 members
Solution:BTGmoderatorLU wrote: ↑Wed Apr 21, 2021 10:00 amSource: Magoosh
Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is the probability that Childeric and Ethelred sit next to each other?
A. 1/30
B. 1/15
C. 1/5
D. 2/5
E. 7/20
The OA is D
We need to determine:
P(Childeric and Ethelred sit next to each other)
This probability is given by (number of arrangements where Childeric and Ethelred sit next to each other) / (total number of ways to arrange the 5 children).
The total number of ways to arrange the 5 children is 5! = 120 ways.
The number of ways with Childeric next to Ethelred can be shown as:
[C-E] - A - B - D
We are treating C and E as a single entity because they must sit next to each other. Thus, we have 4 positions that can be arranged in 4! = 24 ways and we also see that [C-E] can be arranged in 2! = 2 ways, so the total number of arrangements is 24 x 2 = 48.
So, the probability that Childeric and Ethelred sit next to each other is 48/120 = 2/5.
Answer: D
Scott Woodbury-Stewart
Founder and CEO
[email protected]
See why Target Test Prep is rated 5 out of 5 stars on BEAT the GMAT. Read our reviews