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Question Number 133995 by mr W last updated on 26/Feb/21
$${in}\:{how}\:{many}\:{ways}\:{can}\:{n}\:{men}\:{and} \\ $$$${n}\:{women}\:{be}\:{arranged}\:{in}\:{a}\:{row}\:{such} \\ $$$${that}\:{men}\:{and}\:{women}\:{alternate}? \\ $$
Commented by benjo_mathlover last updated on 26/Feb/21
$$=\:\mathrm{2}×\mathrm{n}!×\mathrm{n}!\:=\:\mathrm{2}×\left(\mathrm{n}!\right)^{\mathrm{2}} \\ $$$$\mathrm{example}\:\begin{cases}{\mathrm{3}\:\mathrm{women}}\\{\mathrm{3}\:\mathrm{men}}\end{cases} \\ $$$$\:\mathrm{MWMWMW}\:=\:\mathrm{3}!×\mathrm{3}! \\ $$$$\mathrm{WMWMWM}=\mathrm{3}!×\mathrm{3}! \\ $$$$\mathrm{totally}\:=\:\mathrm{2}×\mathrm{3}!×\mathrm{3}! \\ $$
Commented by mr W last updated on 26/Feb/21
$${thanks}! \\ $$