5 boys and 5 girls randomly select

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NandishSS: Master | Next Rank: 500 Posts; Posts: 366; Joined: Fri Jun 05, 2015 3:35 am; Thanked: 3 times; Followed by:2 members

5 boys and 5 girls randomly select

by NandishSS » Mon Jun 24, 2019 6:49 pm

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5 boys and 5 girls randomly select seats around a circular table that seats 10. What is the probability that two girls will sit next to one another?

A.11/24

B.23/24

C.23/48

D.47/48

E.125/126

OA: E

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Source: Beat The GMAT — Problem Solving | Mon Jun 24, 2019 6:49 pm

regor60: Master | Next Rank: 500 Posts; Posts: 416; Joined: Thu Oct 15, 2009 11:52 am; Thanked: 27 times

by regor60 » Tue Jun 25, 2019 5:57 am

NandishSS wrote:5 boys and 5 girls randomly select seats around a circular table that seats 10. What is the probability that two girls will sit next to one another?

A.11/24

B.23/24

C.23/48

D.47/48

E.125/126

OA: E

The answer works only if the question is interpreted to mean "two or more".

Therefore, to not have any girls sit together means boys and girls must alternate seats. First, pick a seat and place a boy or a girl in it.

Two choices. Then they must alternate.

How many ways are there to seat 10 people around the table ? 10!. But this must be divided by 5!^2 to account for the fact that it doesn't matter which boy or girl since we care only about the gender.

So the probability of no girls (or boys) sitting together is (2)(5!)^2/10! = 1/126

Therefore, the probability of at least two girls sitting together is 1-1/126 = [spoiler] 125/126, E[/spoiler]

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Mon Jul 01, 2019 4:52 pm

NandishSS wrote:5 boys and 5 girls randomly select seats around a circular table that seats 10. What is the probability that two girls will sit next to one another?

A.11/24

B.23/24

C.23/48

D.47/48

E.125/126

OA: E

The only way that two girls will not sit next to one another is if the boys and girls sit alternately next to one another.

10!/(5! x 5!) = (10 x 9 x 8 x 7 x 6)/(5 x 4 x 3 x 2) = 3 x 2 x 7 x 6 = 252

The number of ways BGBGBGBGBG can sit is:

5 x 5 x 4 x 4 x 3 x 3 x 2 x 2 x 1 x 1 = (5!)^2

The number of ways GBGBGBGBGB can sit is also (5!)^2.

Therefore, the probability is:

1 - 2(5!)^2/10! = 1 - 1/126 = 125/126

Answer: E

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