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

(i) the six girls sit together

(ii) the boys and girls sit alternately.

Hello Student,

Please find the answer to your question

(i) 6 boys and 6 girls sit in a row randomly.
Total ways of their seating = 12!
No. of ways in which all the 6 girls sit together = 6! x 7! (Considering all 6 girls as one person)
∴ Probability of all girls sitting together
= 6! x 7!/12! = 720/12 x 11 x 10 x 9 x 8 = 1/132
(ii) Staring with boy, boys can sit in 6! Ways leaving one place between every two boys and two one a last.
B _ B _ B _ B _ B _ B _
These left over places can be occupied by girls in 6! ways.
∴ If we start, with boys. No. of ways of seating boys and girls alternately = 6! x 6!
In the similar manner, if we start with girl, no. of ways of seating boys and girls alternately
= 6! x 6!
G _ G _ G _ G _ G _ G _
Thus total ways of alternate seating arrangements
= 6! x 6! + 6! x 6!
= 2 x 6! x 6!
∴ Probability of making alternate seating arrangement for 6 boys and 6 girls
= 2 x 6! x 6!/12! = 2 x 720/12 x 11 x 10 x 9 x 8 x 7 = 1/462

Thanks
Jitender Pal
askIITians Faculty