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Question Number 103300 by bramlex last updated on 14/Jul/20
$$\mathrm{many}\:\mathrm{positive}\:\mathrm{five}−\mathrm{digit} \\ $$$$\mathrm{integers}\:\mathrm{with}\:\mathrm{the}\:\mathrm{first}\:\mathrm{number}\:\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{there}\:\mathrm{are}\:\mathrm{three}\:\mathrm{equal} \\ $$$$\mathrm{numbers}\:? \\ $$$$\left(\mathrm{a}\right)\:\mathrm{810}\:\:\:\:\left(\mathrm{b}\right)\:\mathrm{720}\:\:\:\:\left(\mathrm{c}\right)\mathrm{120} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{60}\:\:\:\:\:\left(\mathrm{e}\right)\:\mathrm{20} \\ $$
Answered by bemath last updated on 14/Jul/20
$${case}\left(\mathrm{1}\right)\:\mathrm{1\_11}{xx}\:=\mathrm{9}×\frac{\mathrm{4}!}{\mathrm{2}!.\mathrm{2}!}\:=\:\mathrm{54}\: \\ $$$${case}\left(\mathrm{2}\right)\mathrm{1\_11}{xy}\:=\:{C}_{\mathrm{2}} ^{\mathrm{9}} ×\frac{\mathrm{4}!}{\mathrm{2}!}\:=\:\mathrm{432} \\ $$$${case}\left(\mathrm{3}\right)\mathrm{1\_}{xxxy}\:=\:\mathrm{9}×\mathrm{8}×\frac{\mathrm{4}!}{\mathrm{3}!}\:=\:\mathrm{288} \\ $$$${case}\left(\mathrm{4}\right)\mathrm{1\_1}{xxx}=\:\mathrm{9}×\frac{\mathrm{4}!}{\mathrm{3}!}=\mathrm{36} \\ $$$${totally}\:=\:\mathrm{54}+\mathrm{432}+\mathrm{288}+\mathrm{36}\:=\:\mathrm{810} \\ $$
Commented by bemath last updated on 14/Jul/20
$${thank}\:{you}\: \\ $$
Commented by john santu last updated on 14/Jul/20
$${correction}\: \\ $$$${case}\left(\mathrm{3}\right)\:\mathrm{1\_}{xxxy}\:=\:\mathrm{9}×\mathrm{8}×\frac{\mathrm{4}!}{\mathrm{3}!}\:= \\ $$$$\mathrm{288}\:. \\ $$$${totally}\:=\:\mathrm{54}+\mathrm{432}+\mathrm{288}+\mathrm{36} \\ $$$$=\mathrm{810}.\:\left({answer}\:{A}\right)\:\left({JS}\:\circledast\right)\: \\ $$