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Question Number 130034 by EDWIN88 last updated on 22/Jan/21

$$\mathrm{How}\mathrm{many}\mathrm{integers}\mathrm{between}999\mathrm{and}4000$$$$\left(\mathrm{both}\mathrm{are}\mathrm{included}\right)\mathrm{can}\mathrm{be}\mathrm{formed}\mathrm{using}$$$$\mathrm{digits}0,1,2,3,4\mathrm{if}\mathrm{repetition}\mathrm{of}\mathrm{digits}\mathrm{is}$$$$\mathrm{allowed}?$$

Answered by liberty last updated on 22/Jan/21

$$\mathrm{case}\left(1\right)1\mathrm{xyz}=5^3\mathrm{numbers}=125\mathrm{numbers}$$$$\left(2\right)2\mathrm{xyz}=125\mathrm{numbers}$$$$\left(3\right)3\mathrm{xyz}=125\mathrm{numbers}$$$$\left(4\right)4000=1\mathrm{number}$$$$\mathrm{totally}=376\mathrm{numbers}$$

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