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Six boys and six girls are to sit in a row at random. The probability that boys and girls sit alternately is
$
  A.{\text{ }}\dfrac{{6! \times {}^7{P_6}}}{{12!}} \\
  B.{\text{ }}\dfrac{{6!6!}}{{12!}} \\
  C.{\text{ }}\dfrac{{2 \times 6!6!}}{{12!}} \\
  D.{\text{ }}\dfrac{{2 \times 6!}}{{12!}} \\
$

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Hint: First find out how many combinations of boys and girls sitting alternately are possible. Out of all the possible combinations girls would have 6! Ways because no. of girls are 6. But boys will have 7! (by using the concept of permutation) ways because the last boy can have two places either before the first girl or after the last girl and then accordingly calculate the probability.

Complete step-by-step answer:
Total ways in which all the 12 people sit : 12!
$n\left( s \right) = 12!$ ways where S is sample space.
E= Event of the girls and boys to be seated alternately.
6 girls can be seated in 6! ways.
${G_1},{G_2},{G_3},{G_{_4}},{G_5},{G_6}$
There are 7 alternate places for 6 boys.
6 boys can be seated in 7 places in ${}^7{P_6}$ ways.
$\therefore $ $n\left( E \right) = {\text{ 6! }} \times {}^7{{\text{P}}_6}$
$\therefore $ The required probability = $\dfrac{{n\left( E \right)}}{{n\left( S \right)}}$= \[\dfrac{{6!\; \times {}^7{P_6}}}{{12!}}\]
The correct option is A.

Note: The formula for calculating the permutation is given by \[P(n,r) = \dfrac{{n!}}{{(n - r)!}}\], where n is the total no. of object in the set, r is the no. of choosing object from the set.
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