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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 : All the girls are never together

$\begin{array}{1 1}(A)\;10!-5!\times 6!\\(B)\;5!\times 2!\\(C)\;10!-4!\times 5!\\(D)\;\text{None of these}\end{array} $

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The no of ways in which girls are never together =Total no of ways of arrangement - Total no of arrangements in which all the girls together
$\Rightarrow 10!-5!\times 6!$
Hence (A) is the correct answer.
answered May 19, 2014 by sreemathi.v

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