Six boys and six girls sit in a row at random, then the probability that six girls sit together is 1/k, then the value of k is_______.

n(S) =number of ways in which 6 boys and 6 girls can be arrnaged=12!

Required ways=consider six girls as one unit. 6 boys along with the unit of girls can be arranged in 7! ways. 6 girls can be arranged in 6! ways.

n(E)= $=7!6!$
$\because$Required probability $=\frac{7!6!}{12!}=\frac{6!}{12\times 11\times 10\times 9\times 8}=\frac{1}{11\times 12}=\frac{1}{132}$