## 5 boys and 5 girls randomly select

##### This topic has expert replies
Master | Next Rank: 500 Posts
Posts: 367
Joined: 05 Jun 2015
Thanked: 3 times
Followed by:2 members

### 5 boys and 5 girls randomly select

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

00:00

A

B

C

D

E

## Global Stats

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

Master | Next Rank: 500 Posts
Posts: 337
Joined: 15 Oct 2009
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]

### GMAT/MBA Expert

GMAT Instructor
Posts: 6106
Joined: 25 Apr 2015
Location: Los Angeles, CA
Thanked: 43 times
Followed by:24 members
by [email protected] » 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