Question Number 153573 by otchereabdullai@gmail.com last updated on 08/Sep/21
$$\mathrm{A}\:\mathrm{commitee}\:\mathrm{of}\:\mathrm{5}\:\mathrm{men}\:\mathrm{and}\:\mathrm{3}\:\mathrm{women}\:\mathrm{is}\: \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{selected}\:\mathrm{from}\:\mathrm{10}\:\mathrm{men}\:\mathrm{and}\:\mathrm{8}\:\mathrm{women}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{this}\:\mathrm{be}\:\mathrm{done}\:\mathrm{if} \\ $$$$\mathrm{a}\:\mathrm{particular}\:\mathrm{women}\:\mathrm{must}\:\mathrm{be}\:\mathrm{in}\:\mathrm{the}\: \\ $$$$\mathrm{committee} \\ $$
Answered by TheSupreme last updated on 08/Sep/21
$$\mathrm{5}\:{man}\:{from}\:\mathrm{10}\:\left(\frac{\mathrm{10}!}{\mathrm{5}!\mathrm{5}!}\right) \\ $$$$\mathrm{1}\:{womam}\:{selected}\:\mathrm{1} \\ $$$$\mathrm{2}\:{women}\:{from}\:\mathrm{7}\:\left(\frac{\mathrm{7}!}{\mathrm{2}!\mathrm{5}!}\right) \\ $$$${total}\:{ways}\:\frac{\mathrm{10}!\mathrm{7}!}{\mathrm{5}!\mathrm{5}!\mathrm{5}!\mathrm{2}!}=\mathrm{5292} \\ $$$$ \\ $$
Commented by otchereabdullai@gmail.com last updated on 08/Sep/21
$$\mathrm{Am}\:\mathrm{much}\:\mathrm{grateful}\:\mathrm{sir}\:! \\ $$