How-many-6-digits-positive-integers-which-are-formed-by-the-digits-1-to-9-are-such-that-each-of-the-digits-in-the-number-appears-at-least-twice-For-instance-121233-122221-777777-and-etc-

Question Number 116184 by ZiYangLee last updated on 01/Oct/20

$$\mathrm{How}\:\mathrm{many}\:\mathrm{6}-\mathrm{digits}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{which}\:\mathrm{are} \\ $$$$\mathrm{formed}\:\mathrm{by}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{1}\:\mathrm{to}\:\mathrm{9}\:\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\mathrm{each} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{in}\:\mathrm{the}\:\mathrm{number}\:\mathrm{appears}\:\mathrm{at}\:\mathrm{least} \\ $$$$\mathrm{twice}?\:\left[\mathrm{For}\:\mathrm{instance}:\:\mathrm{121233},\mathrm{122221},\mathrm{777777}\:\mathrm{and}\:\mathrm{etc}.\right] \\ $$

Answered by mr W last updated on 01/Oct/20

$${we}\:{have}\:{following}\:{cases}: \\ $$$${AABBCC} \\ $$$$\Rightarrow{C}_{\mathrm{3}} ^{\mathrm{9}} ×\frac{\mathrm{6}!}{\mathrm{2}!\mathrm{2}!\mathrm{2}!}=\mathrm{7560} \\ $$$${AAABBB} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{9}} ×\frac{\mathrm{6}!}{\mathrm{3}!\mathrm{3}!}=\mathrm{720} \\ $$$${AAAABB} \\ $$$$\Rightarrow{C}_{\mathrm{2}} ^{\mathrm{9}} ×\frac{\mathrm{6}!}{\mathrm{4}!\mathrm{2}!}=\mathrm{540} \\ $$$${AAAAAA} \\ $$$$\Rightarrow{C}_{\mathrm{1}} ^{\mathrm{9}} =\mathrm{9} \\ $$$$ \\ $$$${totally}\:\mathrm{7560}+\mathrm{720}+\mathrm{540}+\mathrm{9}=\mathrm{8829} \\ $$

Commented by ZiYangLee last updated on 02/Oct/20

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