6 boys and 6 girls are sitting in a row randomly. The probability that

1.	6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is(a) \(\frac{1}{36}\)(b) \(\frac{1}{462}\)(c) \(\frac{5}{126}\)(d) \(\frac{1}{231}\)
Answer» (b) \(\frac{1}{462}\) Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the 6 girls and 6 boys can be arranged in alternate position in two ways. Ist way : We start with a boy. Then the arrangement is : B G B G B G B G B G B G ∴ Number of ways of arranging 6 boys in 6 places = 6! Number of ways of arranging 6 girls in 6 places = 6! ∴ Number of ways of arranging 6 boys and 6 girls in alternate places = 6! × 6! Similarly, IInd way : Here we start with a girl. Then the arrangement is G B G B G B G B G B G B ∴ Number of ways of arranging 6 boys and 6 girls alternately this way = 6! × 6! ∴ n(E) = 6! × 6! + 6! × 6! = 2 × 6! × 6! ∴ n(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2\times6!\times6!}{12!}\) = \(\frac{2\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7}\) = \(\frac{1}{462}\).

6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is(a) \(\frac{1}{36}\)(b) \(\frac{1}{462}\)(c) \(\frac{5}{126}\)(d) \(\frac{1}{231}\)

Answer»

(b) \(\frac{1}{462}\)

Let S be the sample space.

Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12!

Let E : Event of 6 girls and 6 boys sitting alternately.

Then, the 6 girls and 6 boys can be arranged in alternate position in two ways.

Ist way : We start with a boy. Then the arrangement is : B G B G B G B G B G B G

∴ Number of ways of arranging 6 boys in 6 places = 6!

Number of ways of arranging 6 girls in 6 places = 6!

∴ Number of ways of arranging 6 boys and 6 girls in alternate places = 6! × 6!

Similarly,

IInd way : Here we start with a girl. Then the arrangement is G B G B G B G B G B G B

∴ Number of ways of arranging 6 boys and 6 girls alternately this way

= 6! × 6!

∴ n(E) = 6! × 6! + 6! × 6!

= 2 × 6! × 6!

∴ n(E) = \(\frac{n(E)}{n(S)}\) = \(\frac{2\times6!\times6!}{12!}\)

= \(\frac{2\times6\times5\times4\times3\times2\times1}{12\times11\times10\times9\times8\times7}\) = \(\frac{1}{462}\).