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Suppose 5 boys and 3 girls are to be arranged such that only one girl has a boy adjacent to her . Find the number such arrangement .

My work : I think there are only such possible arrangment $2 \times 3!\times 5! $ .

If the girls are aren't together then one girl will have two boys adjacent to her .So girls must be to-gather and they should be at the either of the two extreme side of the arrangement of the boys . There are two possible side that one of the $3$ girls can take . Now the girls can arrange among themselves in $3!$ ways and boys can arrange among themselves in $5!$ ways.So there are about $3!\times 5! $ arrangements when girls are together and are at the left of the $5$-boys . Similarly there are $3!\times 5!$ other cases when girls are to be put on the left . So total cases= $2\times 3! \times 5! $

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    $\begingroup$ Assuming we are arranging them in a line, yes that sounds good to me. Being a nitpicker though, we could arrange some of the girls on this side of town and the boys on the other side of town in several different more ways. $\endgroup$
    – JMoravitz
    Commented May 16, 2017 at 21:31

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community wiki post so that the question can be closed

Assuming the boys and girls are arranged in a line, your answer is correct.

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