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Question Number 57909 by mr W last updated on 14/Apr/19
$${n}\:{men}\:{and}\:{n}\:{women}\:{should}\:{be}\:{arranged} \\ $$$${alternately}\:{in}\:{a}\:{row},\:{how}\:{many}\:{ways} \\ $$$${can}\:{this}\:{be}\:{done}?\:{if}\:{the}\:{same}\:{should} \\ $$$${be}\:{done}\:{on}\:{a}\:{table},\:{how}\:{many}\:{ways}\:{then}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Apr/19
![trying to understand... n men can be arranged in n! ways now when n men seated ...seats available for women are (n−1)+2←[this 2 seat extreme left and extreme right seat] =n+1 seats so women can be arrsnged n+1_P_n so answer is n!×(n+1_p_n ) =n!×(((n+1)!)/((n+1−n)!)) =n!×(n+1)! Round table... in round table at first one man seat to fixed so n men canbe seated=(n−1)! ways next n women can be seated in available n seat n_p_n =((n!)/((n−n)!))=n! so answer is (n−1)!×n!](Q57928.png)
$${trying}\:{to}\:{understand}... \\ $$$${n}\:{men}\:{can}\:{be}\:{arranged}\:{in}\:{n}!\:{ways} \\ $$$${now}\:{when}\:{n}\:{men}\:{seated}\:...{seats}\:{available}\:{for}\: \\ $$$${women}\:{are}\:\left({n}−\mathrm{1}\right)+\mathrm{2}\leftarrow\left[{this}\:\mathrm{2}\:{seat}\:{extreme}\:{left}\:{and}\:{extreme}\:{right}\:{seat}\right] \\ $$$$={n}+\mathrm{1}\:{seats} \\ $$$${so}\:{women}\:{can}\:{be}\:{arrsnged}\:\:{n}+\mathrm{1}_{{P}_{{n}} } \\ $$$${so}\:{answer}\:{is}\:\:\:\boldsymbol{{n}}!×\left(\boldsymbol{{n}}+\mathrm{1}_{\boldsymbol{{p}}_{\boldsymbol{{n}}} } \right) \\ $$$$={n}!×\frac{\left({n}+\mathrm{1}\right)!}{\left({n}+\mathrm{1}−{n}\right)!} \\ $$$$={n}!×\left({n}+\mathrm{1}\right)! \\ $$$${Round}\:{table}... \\ $$$${in}\:{round}\:{table}\:{at}\:{first}\:{one}\:{man}\:{seat}\:{to}\:{fixed} \\ $$$${so}\:{n}\:{men}\:{canbe}\:{seated}=\left({n}−\mathrm{1}\right)!\:{ways} \\ $$$${next}\:{n}\:{women}\:{can}\:{be}\:{seated}\:{in}\:{available}\:{n}\:{seat} \\ $$$${n}_{{p}_{{n}} } =\frac{{n}!}{\left({n}−{n}\right)!}={n}! \\ $$$${so}\:{answer}\:{is}\:\left({n}−\mathrm{1}\right)!×{n}! \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 14/Apr/19
$${please}\:{check}\:{sir}: \\ $$$${part}\:\mathrm{1}: \\ $$$${n}+\mathrm{1}\:{seats}\:{for}\:{women},\:{that}'{s}\:{correct}. \\ $$$${but}\:{not}\:{all}\:{possibilities}\:{to}\:{occupy} \\ $$$${these}\:{seats}\:{are}\:{allowed},\:{e}.{g}. \\ $$$${WMWMWMWM}...{MMWMW} \\ $$$${because}\:{men}\:{and}\:{women}\:{must}\:{sit} \\ $$$${alternately}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 14/Apr/19
$${thank}\:{you}\:{sir}...{for}\:{rectification} \\ $$
Answered by mr W last updated on 14/Apr/19
$${in}\:{a}\:{row}: \\ $$$${to}\:{arrange}\:{n}\:{men}\:{there}\:{are}\:{n}!\:{ways}. \\ $$$${to}\:{arrange}\:{n}\:{women}: \\ $$$${either}\:{WMWMWM}...{WMWMWM}\:\Rightarrow{n}!\:{ways} \\ $$$${or}\:{MWMWMW}...{MWMWMW}\:\Rightarrow{n}!\:{ways} \\ $$$${totally}\:\mathrm{2}{n}!{n}!=\mathrm{2}\left({n}!\right)^{\mathrm{2}} \:{ways} \\ $$$$ \\ $$$${on}\:{a}\:{table}: \\ $$$${to}\:{arrange}\:{n}\:{men}\:{there}\:{are}\:\left({n}−\mathrm{1}\right)!\:{ways}. \\ $$$${to}\:{arrange}\:{n}\:{women}\:{there}\:{are}\:{n}!\:{ways}. \\ $$$${totally}\:{n}!\left({n}−\mathrm{1}\right)! \\ $$