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-

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

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

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

$$wehavefollowingcases:$$$$AABBCC$$$$\Rightarrow C_3^9\times \frac{6!}{2!2!2!}=7560$$$$AAABBB$$$$\Rightarrow C_2^9\times \frac{6!}{3!3!}=720$$$$AAAABB$$$$\Rightarrow C_2^9\times \frac{6!}{4!2!}=540$$$$AAAAAA$$$$\Rightarrow C_1^9=9$$$$$$$$totally7560+720+540+9=8829$$

Commented by ZiYangLee last updated on 02/Oct/20

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