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Five boys and five girls form a line.Find the number of ways of making the seating arrangement under the following condition : Boys and girls alternate

$\begin{array}{1 1}(A)\;(5!)^2+(5!)^2\\(B)\;(2!)^2+(2!)^2\\(C)\;(3!)^2+(3!)^2\\(D)\;\text{None of these}\end{array} $

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No of ways the girls can sit =5!
No of ways the boys can sit =5!
Boys and girls alternate =$5!\times 5!$
Vice versa =$5!\times 5!$
$\Rightarrow (5!)^2+(5!)^2$
Hence (A) is the correct answer.
answered May 19, 2014 by sreemathi.v

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