Browse Questions

Five boys and five girls form a line.Find the number of ways of making the seating arrangement under the following condition : No two girls sit together

$\begin{array}{1 1}(A)\;5!\times 6!\\(B)\;4!\times 6!\\(C)\;3!\times 6!\\(D)\;1!\times 5!\end{array}$

5 boys can be seated in a row in =$5P_5=5!$ ways
Now,in the 6 gaps 5 girls can be arranged in =$6P_5$ ways
Hence the no of ways in which no two girls sit together =$5!\times 6P_5$
$\Rightarrow 5!\times 6!$
Hence (A) is the correct answer.