Question Number 117866 by aurpeyz last updated on 14/Oct/20
$${how}\:{many}\:{four}\:{digit}\:{numbers}\:{can}\:{be} \\ $$$${formed}\:{with}\:{the}\:{digits}\:\mathrm{0}\:\mathrm{1}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{5}\:\mathrm{6}\:\mathrm{7}\:\mathrm{8} \\ $$$$\mathrm{9}? \\ $$
Commented by mr W last updated on 14/Oct/20
$${with}\:{repetition}: \\ $$$$\mathrm{1000},\mathrm{1001},…,\mathrm{9999} \\ $$$$\Rightarrow\mathrm{9999}−\mathrm{999}=\mathrm{9000}\:{numbers} \\ $$
Commented by mhmoud last updated on 14/Oct/20
good
Commented by mr W last updated on 14/Oct/20
$${without}\:{repetition}: \\ $$$$\frac{\mathrm{9}}{\mathrm{10}}×{P}_{\mathrm{4}} ^{\mathrm{10}} =\mathrm{4536}\:{numbers} \\ $$$${or} \\ $$$${P}_{\mathrm{4}} ^{\mathrm{10}} −{P}_{\mathrm{3}} ^{\mathrm{9}} =\mathrm{4536}\:{numbers} \\ $$
Answered by 1549442205PVT last updated on 14/Oct/20
$$\mathrm{10}^{\mathrm{4}} −\mid\overline {\mathrm{0abc}}\mid=\mathrm{10}^{\mathrm{4}} −\mathrm{10}^{\mathrm{3}} =\mathrm{10000}−\mathrm{1000} \\ $$$$=\mathrm{9000} \\ $$
Answered by bobhans last updated on 14/Oct/20
$$\mathrm{9}×\mathrm{10}^{\mathrm{3}} \:=\mathrm{9},\mathrm{000} \\ $$