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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 sit together

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

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A)
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Two groups of girls and boys can be arranged among themselves in 5!.Similarly boys can be arranged in 5!
Total no of ways =$2!(5!\times 5!)$
Hence (B) is the correct answer.
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