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Question Number 199031 by cortano12 last updated on 27/Oct/23
Answered by mr W last updated on 27/Oct/23
$$\left.\mathrm{1}\right) \\ $$$${C}_{\mathrm{3}} ^{\mathrm{4}} ×\mathrm{3}!×{C}_{\mathrm{2}} ^{\mathrm{5}} ×\mathrm{2}!=\mathrm{4}×\mathrm{6}×\mathrm{10}×\mathrm{2}=\mathrm{480} \\ $$$$\left.\mathrm{2}\right) \\ $$$$\frac{\mathrm{4}}{\mathrm{6}}×{C}_{\mathrm{4}} ^{\mathrm{6}} ×\mathrm{4}!−{C}_{\mathrm{2}} ^{\mathrm{4}} ×\mathrm{2}!=\mathrm{228} \\ $$$$\left.\mathrm{3}\right) \\ $$$${all}:\:\mathrm{9}!=\mathrm{362880} \\ $$$${divisible}\:{by}\:\mathrm{4}:\:\mathrm{16}×\mathrm{7}!=\mathrm{80640} \\ $$$$\left.\mathrm{4}\right) \\ $$$$\mathrm{6}!×\mathrm{3}!=\mathrm{4320} \\ $$
Commented by cortano12 last updated on 27/Oct/23
$$\mathrm{explain}\:\mathrm{sir}\:\mathrm{why}\:\mathrm{16}×\mathrm{7}!\: \\ $$
Commented by mr W last updated on 27/Oct/23
$${abcdefgxy} \\ $$$${such}\:{that}\:{this}\:{number}\:{is}\:{divisible} \\ $$$${by}\:\mathrm{4},\:{xy}\:{must}\:{be}\:{divisible}\:{by}\:\mathrm{4}.\:{there} \\ $$$${are}\:\mathrm{16}\:{possiblities}\:{for}\:{xy}: \\ $$$$\mathrm{12}/\mathrm{16}/\mathrm{24}/\mathrm{28}/\mathrm{32}/\mathrm{36}/\mathrm{48}/\mathrm{52}/\mathrm{56}/\mathrm{64}/\mathrm{68}/\mathrm{72}/\mathrm{76}/\mathrm{84}/\mathrm{92}/\mathrm{96} \\ $$$${to}\:{arrange}\:{the}\:{remaining}\:\mathrm{7}\:{digits} \\ $$$${abcdefg}\:{there}\:{are}\:\mathrm{7}!\:{ways}.\:{so}\:{totally} \\ $$$${there}\:{are}\:\mathrm{16}×\mathrm{7}!\:{valid}\:{numbers}. \\ $$
Commented by cortano12 last updated on 27/Oct/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\: \\ $$