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

## 5 boys and 5 girls randomly select

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

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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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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