Question Number 27513 by NECx last updated on 08/Jan/18
$${Using}\:{only}\:{the}\:{integers}\:\mathrm{4}\:{to}\:\mathrm{8}, \\ $$$${how}\:{many}\:{even}\:{numbers}\:{can}\:{be} \\ $$$${formed}\:{if}\:{each}\:{must}\:{lie}\:{between} \\ $$$$\mathrm{4000}\:{and}\:\:\mathrm{9000}? \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 08/Jan/18
$$\mathcal{A}{s}\:{the}\:{numbers}\:{are}\:{between} \\ $$$$\mathrm{4000}\:\&\:\mathrm{9000},\:{so}\:{the}\:{number}\:{of} \\ $$$${positions}\:{is}\:\:\mathrm{4}. \\ $$$$\mathcal{E}{ach}\:{of}\:\mathrm{1000}{th},\mathrm{100}{th}\:\&\:\mathrm{10}{th}\: \\ $$$${positions}\:{can}\:{be}\:{filled}\:{with} \\ $$$$\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7}\:\&\:\mathrm{8}\:{i}-{e}\:{in}\:\mathrm{5}\:{ways}\:{but} \\ $$$${the}\:{unit}\:{position}\:{can}\:{be}\:{filled} \\ $$$${with}\:\mathrm{4},\mathrm{6}\:\&\:\mathrm{8}\:{i}-{e}\:{in}\:\mathrm{3}\:{ways}\:{because} \\ $$$${the}\:{numbers}\:{are}\:{even}. \\ $$$$\therefore{Total}\:{ways}\:=\mathrm{5}.\mathrm{5}.\mathrm{5}.\mathrm{3}=\mathrm{375} \\ $$