5
$\begingroup$

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! $

$\endgroup$
  • 1
    $\begingroup$ 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 May 16 '17 at 21:31
1
$\begingroup$

community wiki post so that the question can be closed

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.