Let x denote the number of ways of arranging m boys and m girls in a row so that no two boys sit together. If y and z give the number of ways of arranging m boys and m girls in a row and around a circular table respectively so that boys and girls sit alternatively, then x : y : z =

$$\begin{gathered} \mathrm{x}=\mathrm{m} \mathrm{boys}, \mathrm{m} \mathrm{girls} \mathrm{arranged} \mathrm{in} \mathrm{a} \mathrm{row} \mathrm{so} \mathrm{that} \mathrm{no} \mathrm{two} \mathrm{boys} \mathrm{sit} \mathrm{together}. \\ \mathrm{m} \mathrm{girls} \mathrm{can} \mathrm{be} \mathrm{arranged} \mathrm{in} \mathrm{row} \mathrm{in} \mathrm{m}! \mathrm{ways}. \\ \mathrm{There} \mathrm{are} \mathrm{m}+1 \mathrm{places} \mathrm{to} \mathrm{arrange} \mathrm{m} \mathrm{boys} \mathrm{in} \mathrm{between} \mathrm{the} \mathrm{girls} \\ \therefore \mathrm{x}=\mathrm{m}! {(}^{m+1}{P}_m)=\mathrm{m}!(\mathrm{m}+1)! \end{gathered}$$$$\begin{gathered} \mathrm{y}=\mathrm{m} \mathrm{boys}, \mathrm{m} \mathrm{girls} \mathrm{to} \mathrm{be} \mathrm{arranged} \mathrm{in} \mathrm{a} \mathrm{row} \mathrm{alternately} \\ \boxed{ \mathrm{y}=2(\mathrm{m}!) \mathrm{m}!} \\ \mathrm{z}=\mathrm{m} \mathrm{boys}, \mathrm{m} \mathrm{girls} \mathrm{to} \mathrm{be} \mathrm{arranged} \mathrm{in} \mathrm{a} \mathrm{circle} \mathrm{alternately} \\ \mathrm{Arrange} \mathrm{m} \mathrm{boys} \mathrm{around} \mathrm{a} \mathrm{circle} \mathrm{in} (\mathrm{m}-1)! \mathrm{ways}. \\ \mathrm{There} \mathrm{are} \mathrm{m} \mathrm{places} \mathrm{to} \mathrm{arrange} \mathrm{m} \mathrm{girls} \mathrm{in} \mathrm{between} \mathrm{m} \mathrm{boys} \\ \therefore \mathrm{Girls} \mathrm{can} \mathrm{be} \mathrm{arranged} \mathrm{in} \mathrm{m}! \mathrm{ways}. \\ \Rightarrow \boxed{\mathrm{z}=(\mathrm{m}-!)! \mathrm{m}!} \\ \mathrm{x}:\mathrm{y}:\mathrm{z}=\mathrm{m}!(\mathrm{m}+1)! : 2(\mathrm{m}!) (\mathrm{m}!) : (\mathrm{m}-1)\mathrm{m}! \\ =(\mathrm{m}+1)! : 2\mathrm{m}! : (\mathrm{m}-1)! \\ =(\mathrm{m}+1)\mathrm{m} : 2\mathrm{m} :\hspace{0.17em}1 \end{gathered}$$