Question Number 124237 by mr W last updated on 01/Dec/20
$${How}\:{many}\:\mathrm{6}\:{digit}\:{odd}\:{numbers}\:{have} \\ $$$${different}\:{digits}? \\ $$
Commented by mr W last updated on 02/Dec/20
$$\mathrm{5}×\left({P}_{\mathrm{5}} ^{\mathrm{9}} −{P}_{\mathrm{4}} ^{\mathrm{8}} \right)=\mathrm{67200} \\ $$
Answered by liberty last updated on 01/Dec/20
$${case}\left(\mathrm{1}\right)\:{A}\Box\Box\Box\Box{B}\:{where}\:{A}\:{and}\:{B}\:{odd}\:{number} \\ $$$$\:=\:\mathrm{5}×{P}\:_{\mathrm{4}} ^{\mathrm{8}} \:×\mathrm{4}\: \\ $$$${case}\left(\mathrm{2}\right)\:{A}\Box\Box\Box\Box{B}\:{where}\:{A}\:{even}\:{and}\:{B}\:{odd}\:{number} \\ $$$$=\:\mathrm{4}×\:{P}\:_{\mathrm{4}} ^{\mathrm{8}} ×\mathrm{5}\: \\ $$$${totally}\:=\:{P}\:_{\mathrm{4}} ^{\mathrm{8}} \:×\:\mathrm{40}\:=\:\frac{\mathrm{8}×\mathrm{7}×\mathrm{6}×\mathrm{5}×\mathrm{4}!}{\mathrm{4}!}\:×\mathrm{40} \\ $$$$\:=\:\mathrm{56}×\mathrm{30}×\mathrm{40}=\mathrm{67200} \\ $$$$\mathrm{56}×\mathrm{30}×\mathrm{40} \\ $$$$\mathrm{67200}.\mathrm{0} \\ $$$$ \\ $$
Commented by mr W last updated on 02/Dec/20
$${correct}\:{sir},\:{thanks}! \\ $$