6 boys and 6 girls sit in a row a row at random. The probability that

1.	6 boys and 6 girls sit in a row a row at random. The probability that all the girls sit together isA. \(\frac{1}{432}\) B. \(\frac{1}{431}\) C. \(\frac{1}{432}\) D. none of these
Answer» 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.

6 boys and 6 girls sit in a row a row at random. The probability that all the girls sit together isA. \(\frac{1}{432}\) B. \(\frac{1}{431}\) C. \(\frac{1}{432}\) D. none of these

Answer»

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.

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6 boys and 6 girls sit in a row a row at random. The probability that all the girls sit together isA. \(\frac{1}{432}\) B. \(\frac{1}{431}\) C. \(\frac{1}{432}\) D. none of these

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