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Question Number 207179 by efronzo1 last updated on 08/May/24
Answered by mr W last updated on 08/May/24
$$\mathrm{3}×\mathrm{4}×\mathrm{3}×\mathrm{2}=\mathrm{72}\:{such}\:{numbers} \\ $$
Commented by efronzo1 last updated on 09/May/24
$$\:\cancel{\underbrace{\boldsymbol{{B}}}} \underbrace{ } \\ $$
Commented by mr W last updated on 09/May/24
$${no}\:{sir}! \\ $$$${to}\:{select}\:{the}\:{last}\:{digit}\:{there}\:{are}\:\mathrm{3} \\ $$$${possibilities}\:\left(\mathrm{1},\:\mathrm{3},\:\mathrm{5}\right).\:{for}\:{the}\: \\ $$$${remaining}\:\mathrm{4}\:{digits}\:{there}\:{are}\:\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1} \\ $$$${possibilities}.\:{so}\:{totally}\:{there}\:{are} \\ $$$$\mathrm{3}×\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}=\mathrm{72}\:{numbers}\:{if} \\ $$$${no}\:{repetition}\:{is}\:{allowed}. \\ $$
Answered by som(math1967) last updated on 08/May/24
$$\:\mathrm{5}^{\mathrm{4}} ×\mathrm{3},\:{if}\:{repetition}\:{is}\:{allowed} \\ $$
Commented by som(math1967) last updated on 09/May/24
$${If}\:{repetition}\:{not}\:{allow} \\ $$$$\blacksquare\blacksquare\blacksquare\blacksquare\mathrm{1}{or}\mathrm{3}{or}\mathrm{5} \\ $$$$\:{P}_{\mathrm{4}} ^{\mathrm{4}} ×\mathrm{3}=\mathrm{4}!×\mathrm{3}=\mathrm{72} \\ $$