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Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is

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Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What is

by Gmat_mission » Sun Jun 14, 2020 2:15 pm

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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

[spoiler]OA=D[/spoiler]

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Re: Five children, Anaxagoras, Beatrice, Childeric, Desdemona, and Ethelred, sit randomly in five chairs in a row. What

by Scott@TargetTestPrep » Sat Jun 20, 2020 2:42 pm
Gmat_mission wrote: ↑
Sun Jun 14, 2020 2:15 pm
 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

[spoiler]OA=D[/spoiler]

Solution:

We need to determine:

P(Childeric and Ethelred sit next to each other)

This probability is given by (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 (which are C - E and E - C), 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

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