Question Number 218578 by mr W last updated on 13/Apr/25

$$\mathrm{10}\:{couples}\:{are}\:{invited}\:{to}\:{a}\:{dinner}. \\ $$$${after}\:{the}\:{dinner}\:{they}\:{form}\:{pairs}\: \\ $$$${to}\:{dance}.\:{in}\:{how}\:{many}\:{ways}\:{can}\:{they} \\ $$$${do}\:{this}, \\ $$$$\left.\mathrm{1}\right)\:{generally} \\ $$$$\left.\mathrm{2}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman} \\ $$$$\left.\mathrm{3}\left.\right)\:{as}\:\mathrm{2}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$$$\left.\mathrm{4}\right)\:{a}\:{man}\:{should}\:{dance}\:{with}\:{a}\:{woman}, \\ $$$$\:\:\:\:\:{but}\:{not}\:{with}\:{his}\:{own}\:{wife} \\ $$$$\left.\mathrm{5}\left.\right)\:{as}\:\mathrm{4}\right),\:{but}\:{two}\:{special}\:{couples} \\ $$$$\:\:\:\:\:{should}\:{not}\:{dance}\:{with}\:{each}\:{other} \\ $$
Answered by MrGaster last updated on 13/Apr/25

$$\left.\mathrm{1}\right)\left(\mathrm{2}\centerdot\mathrm{10}−\mathrm{1}\right)!!=\mathrm{19}!! \\ $$$$\left.\mathrm{2}\right)\mathrm{10}! \\ $$$$\left.\mathrm{3}\right)\mathrm{10}!−\mathrm{2}\centerdot\mathrm{9}!+\mathrm{8} \\ $$
Answered by mr W last updated on 18/Apr/25

$$\left.\mathrm{1}\right) \\ $$$${to}\:{divide}\:\mathrm{20}\:{persons}\:{into}\:\mathrm{10}\:{groups} \\ $$$${with}\:\mathrm{2}\:{persons}\:{in}\:{each}\:{group}\:{there} \\ $$$${are}\:{totally}\:\frac{\mathrm{20}!}{\left(\mathrm{2}!\right)^{\mathrm{10}} }\:{ways}. \\ $$$$\left.\mathrm{2}\right)\: \\ $$$$\mathrm{10}!=\mathrm{3}\:\mathrm{628}\:\mathrm{800} \\ $$$$ \\ $$$$\left.\mathrm{3}\right) \\ $$$${total}\:{number}\:{of}\:{ways}:\:\mathrm{10}! \\ $$$${but}\:{among}\:{them},\: \\ $$$$\left(\mathrm{3}.\mathrm{1}\right)\:{H}_{\mathrm{1}} \Leftrightarrow{W}_{\mathrm{2}} \:{and}\:{H}_{\mathrm{2}} \Leftrightarrow{W}_{\mathrm{1}} : \\ $$$$\mathrm{8}!\:{ways} \\ $$$$\left(\mathrm{3}.\mathrm{2}\right)\:{H}_{\mathrm{1}} \Leftrightarrow{W}_{\mathrm{2}} \:{or}\:{H}_{\mathrm{2}} \Leftrightarrow{W}_{\mathrm{1}} : \\ $$$$\mathrm{2}\left(\mathrm{9}!−\mathrm{8}!\right) \\ $$$${number}\:{of}\:{valid}\:{ways}: \\ $$$$\mathrm{10}!−\mathrm{8}!−\mathrm{2}\left(\mathrm{9}!−\mathrm{8}!\right) \\ $$$$=\mathrm{10}!−\mathrm{2}×\mathrm{9}!+\mathrm{8}!=\mathrm{2}\:\mathrm{943}\:\mathrm{360} \\ $$$$ \\ $$$$\left.\mathrm{4}\right) \\ $$$$!\mathrm{10}=\mathrm{1}\:\mathrm{334}\:\mathrm{961} \\ $$$$ \\ $$$$\left.\mathrm{5}\right) \\ $$$${H}_{{i}} \:{not}\:{with}\:{W}_{{i}} ,\:{i}=\mathrm{1},\mathrm{2},\mathrm{3},\:…,\:\mathrm{10} \\ $$$${H}_{\mathrm{1}} \:{not}\:{with}\:{W}_{\mathrm{2}} \\ $$$${H}_{\mathrm{2}} \:{not}\:{with}\:{W}_{\mathrm{1}} \\ $$$$\blacksquare\blacksquare\Box\Box\Box\Box\Box\Box\Box\Box \\ $$$$\blacksquare\blacksquare\Box\Box\Box\Box\Box\Box\Box\Box \\ $$$$\Box\Box\blacksquare\Box\Box\Box\Box\Box\Box\Box \\ $$$$\Box\Box\Box\blacksquare\Box\Box\Box\Box\Box\Box \\ $$$$\Box\Box\Box\Box\blacksquare\Box\Box\Box\Box\Box \\ $$$$\Box\Box\Box\Box\Box\blacksquare\Box\Box\Box\Box \\ $$$$\Box\Box\Box\Box\Box\Box\blacksquare\Box\Box\Box \\ $$$$\Box\Box\Box\Box\Box\Box\Box\blacksquare\Box\Box \\ $$$$\Box\Box\Box\Box\Box\Box\Box\Box\blacksquare\Box \\ $$$$\Box\Box\Box\Box\Box\Box\Box\Box\Box\blacksquare \\ $$$${there}\:{are}\:\mathrm{1}\:\mathrm{053}\:\mathrm{136}\:{ways}. \\ $$
Commented by mr W last updated on 18/Apr/25
