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Question Number 17770 by chux last updated on 10/Jul/17

$$\mathrm{Find}\mathrm{the}\mathrm{number}\mathrm{of}\mathrm{ways}\mathrm{the}$$$$\mathrm{digits}0,1,2\mathrm{and}3\mathrm{can}\mathrm{be}\mathrm{permuted}$$$$\mathrm{to}\mathrm{give}\mathrm{rise}\mathrm{to}\mathrm{a}\mathrm{number}\mathrm{greater}$$$$\mathrm{than}2000.$$

Answered by alex041103 last updated on 10/Jul/17

$$\mathrm{So}\mathrm{we}\mathrm{search}\mathrm{for}\mathrm{numbers}\overline{\mathrm{abcd}}>2000$$$$\mathrm{and}\mathrm{a}\in \{0,1,2,3\},\mathrm{b}\in \{0,1,2,3\}/\left\{\mathrm{a}\right\},$$$$\mathrm{c}\in \{0,1,2,3\}/\{\mathrm{a},\mathrm{b}\}\mathrm{and}\mathrm{d}\in \{0,1,2,3\}/\{\mathrm{a},\mathrm{b},\mathrm{c}\}$$$$\mathrm{Also}\mathrm{in}\mathrm{order}\mathrm{for}\overline{\mathrm{abcd}}>2000\to \mathrm{a}\geqq 2$$$$\mathrm{So}\mathrm{for}\mathrm{the}\mathrm{first}\mathrm{digit}\mathrm{we}\mathrm{have}2\mathrm{possabilities}.$$$$\mathrm{For}\mathrm{the}\mathrm{second},\mathrm{third}\mathrm{and}\mathrm{fourth}\mathrm{digit}$$$$\mathrm{we}\mathrm{have}3!=6\mathrm{possable}\mathrm{permutations}.$$$$\mathrm{Even}\mathrm{the}\mathrm{smallest}\overline{\mathrm{abcd}}\mathrm{we}\mathrm{can}\mathrm{have}\mathrm{where}$$$$\mathrm{a}\geqq 2,\mathrm{i}.\mathrm{e}.2013\mathrm{is}\mathrm{bigger}\mathrm{than}2000.\mathrm{Thats}$$$$\mathrm{why}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{restrictionsfor}\mathrm{the}\mathrm{last}$$$$\mathrm{three}\mathrm{digits}.$$$$\mathrm{Total}:2\times 3!=2\times 6=12$$

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