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Question Number 17770 by chux last updated on 10/Jul/17
Find the number of ways the digits 0,1,2 and 3 can be permuted to give rise to a number greater than 2000.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{ways}\:\mathrm{the} \\ $$$$\mathrm{digits}\:\mathrm{0},\mathrm{1},\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{can}\:\mathrm{be}\:\mathrm{permuted} \\ $$$$\mathrm{to}\:\mathrm{give}\:\mathrm{rise}\:\mathrm{to}\:\mathrm{a}\:\mathrm{number}\:\mathrm{greater} \\ $$$$\mathrm{than}\:\mathrm{2000}. \\ $$
Answered by alex041103 last updated on 10/Jul/17
So we search for numbers abcd^(−) >2000 and a∈{0,1,2,3}, b∈{0,1,2,3}/{a}, c∈{0,1,2,3}/{a,b} and d∈{0,1,2,3}/{a,b,c} Also in order for abcd^(−) >2000 →a≧2 So for the first digit we have 2 possabilities. For the second, third and fourth digit we have 3!=6 possable permutations. Even the smallest abcd^(−) we can have where a≧2, i.e. 2013 is bigger than 2000. Thats why we don′t have restrictionsfor the last three digits. Total: 2×3!=2×6=12
$$\mathrm{So}\:\mathrm{we}\:\mathrm{search}\:\mathrm{for}\:\mathrm{numbers}\:\overline {\mathrm{abcd}}>\mathrm{2000} \\ $$$$\mathrm{and}\:\mathrm{a}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\},\:\mathrm{b}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}/\left\{\mathrm{a}\right\}, \\ $$$$\mathrm{c}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}/\left\{\mathrm{a},\mathrm{b}\right\}\:\mathrm{and}\:\mathrm{d}\in\left\{\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3}\right\}/\left\{\mathrm{a},\mathrm{b},\mathrm{c}\right\} \\ $$$$\mathrm{Also}\:\mathrm{in}\:\mathrm{order}\:\mathrm{for}\:\overline {\mathrm{abcd}}>\mathrm{2000}\:\rightarrow\mathrm{a}\geqq\mathrm{2} \\ $$$$\mathrm{So}\:\mathrm{for}\:\mathrm{the}\:\mathrm{first}\:\mathrm{digit}\:\mathrm{we}\:\mathrm{have}\:\mathrm{2}\:\mathrm{possabilities}. \\ $$$$\mathrm{For}\:\mathrm{the}\:\mathrm{second},\:\mathrm{third}\:\mathrm{and}\:\mathrm{fourth}\:\mathrm{digit} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{3}!=\mathrm{6}\:\mathrm{possable}\:\mathrm{permutations}. \\ $$$$\mathrm{Even}\:\mathrm{the}\:\mathrm{smallest}\:\overline {\mathrm{abcd}}\:\mathrm{we}\:\mathrm{can}\:\mathrm{have}\:\mathrm{where} \\ $$$$\mathrm{a}\geqq\mathrm{2},\:\mathrm{i}.\mathrm{e}.\:\mathrm{2013}\:\mathrm{is}\:\mathrm{bigger}\:\mathrm{than}\:\mathrm{2000}.\:\mathrm{Thats} \\ $$$$\mathrm{why}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{have}\:\mathrm{restrictionsfor}\:\mathrm{the}\:\mathrm{last} \\ $$$$\mathrm{three}\:\mathrm{digits}. \\ $$$$\mathrm{Total}:\:\mathrm{2}×\mathrm{3}!=\mathrm{2}×\mathrm{6}=\mathrm{12} \\ $$

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