Question

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

A.  \(\frac{1}{432}\) 

B. \(\frac{1}{431}\) 

C.  \(\frac{1}{432}\) 

D. none of these

Answer 1

As 6 boys and 6 girls are sitting in a row. So these 12 persons can sit in 12! Ways 

Now group all 6 girls together and treat them as 1. 

Now, all girls together can sit in 7! Ways 

And girls can sit among self in 6! Ways. 

∴ total ways in which all 6 girls sit together = 7! × 6!

∴ P(E) = \(\frac{7!\times 6!}{12!}\) = \(\frac{720}{8\times9\times10\times11\times12}\) = \(\frac{1}{132}\)

Hence,  

P(E) = \(\frac{1}{132}\)

As our answer matches only with option (d) 

∴ Option (d) is the only correct choice.