14.Probability
medium

Six boys and six girls sit in a row. What is the probability that the boys and girls sit alternatively

A

$\frac{1}{{462}}$

B

$\frac{1}{{924}}$

C

$\frac{1}{2}$

D

None of these

(IIT-1979)

Solution

(a) Let $n = $ total number of ways $ = 12\,!$

and $m = $ favourable number of ways $ = 2 \times 6\,!\,.\,6\,!$

Since the boys and girls can sit alternately in $6\,!\,\,.\,\,6\,!$ ways if we begin with a

boy and similarly they can sit alternately in $6\,!\,\,.\,\,6\,!$ ways if we begin with a girl

Hence required probability $ = \frac{m}{n} = \frac{{2 \times 6\,!\,\,.\,\,6\,!}}{{12\,!}} = \frac{1}{{462}}$.

Standard 11
Mathematics

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