Question

Six boys and girls sit in a row randomly. Find the probability that the six girls sit together.

• A. $p=\frac{1}{95040}$
• B. $p=\frac{1}{132}$
• C. $p=\frac{1}{924}$
• D. $p=\frac{1}{66}$

Answer 1

Answer: (B)

$p=\frac{1}{132}$

Total number of boys and girls $=6+6=12$
Total cases $=12!$
Since, six girls are to be seated together , so they can be considered as 1 object .
So, total number of persons to be arranged will be 6 boys and 1 group.
And the six girls can arrange themselves in $6!$ ways.
So, favorable cases $=7!6!$
The required probability $=\frac{7!6!}{12!}=\frac{1}{132}$