Question

6 boys and 6 girls sit in a row at random. The probability that all the girls sit together is

Options

• \[\frac{1}{432}\]

   

• \[\frac{12}{431}\]

   

• \[\frac{1}{132}\]

   

•  None of these      

Answer 1

Total number of ways in which 6 boys and 6 girls can sit in a row = 12!
Consider 6 girls as one group, then 6 boys and one group can arrange in 7! ways.
Now, 6 girls in the group can arrange among themselves in 6!.
So, the number of ways in which all the girls sit together is 7! × 6!.
∴ P(all girls sit together) = \[\frac{\text{ Number of ways in which all girls sit together} }{\text{ Total number of ways in which 6 boys and 6 girls sit in a row} } = \frac{7! 6!}{12!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{12 \times 11 \times 10 \times 9 \times 8} = \frac{1}{132}\]