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
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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.