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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