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Find-the-number-of-odd-integers-between-30-000-and-80-000-in-which-no-digit-is-repeated-




Question Number 18655 by Tinkutara last updated on 26/Jul/17
Find the number of odd integers  between 30,000 and 80,000 in which no  digit is repeated.
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{odd}\:\mathrm{integers} \\ $$$$\mathrm{between}\:\mathrm{30},\mathrm{000}\:\mathrm{and}\:\mathrm{80},\mathrm{000}\:\mathrm{in}\:\mathrm{which}\:\mathrm{no} \\ $$$$\mathrm{digit}\:\mathrm{is}\:\mathrm{repeated}. \\ $$
Commented by mrW1 last updated on 27/Jul/17
let′s say the number is XZZZY  X can be 3,4,5,6,7  Y can be 1,3,5,7,9  Z can be 0,1,2,3,4,5,6,7,8,9  case 1:  Y is one from 3 digits in {3,5,7}  X can be one from 4 digits: 4,6 and 2 from {3,5,7}  ZZZ can be choosen from 8 digits  ⇒3×4×8×7×6  case 2:  Y is one from 2 digits in {1,9}  X can be one from 5 digits in [3,7]  ZZZ can be choosen from 8 digits  ⇒2×5×8×7×6    totally  ⇒3×4×8×7×6+2×5×8×7×6=7392
$$\mathrm{let}'\mathrm{s}\:\mathrm{say}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{XZZZY} \\ $$$$\mathrm{X}\:\mathrm{can}\:\mathrm{be}\:\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7} \\ $$$$\mathrm{Y}\:\mathrm{can}\:\mathrm{be}\:\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7},\mathrm{9} \\ $$$$\mathrm{Z}\:\mathrm{can}\:\mathrm{be}\:\mathrm{0},\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6},\mathrm{7},\mathrm{8},\mathrm{9} \\ $$$$\mathrm{case}\:\mathrm{1}: \\ $$$$\mathrm{Y}\:\mathrm{is}\:\mathrm{one}\:\mathrm{from}\:\mathrm{3}\:\mathrm{digits}\:\mathrm{in}\:\left\{\mathrm{3},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{X}\:\mathrm{can}\:\mathrm{be}\:\mathrm{one}\:\mathrm{from}\:\mathrm{4}\:\mathrm{digits}:\:\mathrm{4},\mathrm{6}\:\mathrm{and}\:\mathrm{2}\:\mathrm{from}\:\left\{\mathrm{3},\mathrm{5},\mathrm{7}\right\} \\ $$$$\mathrm{ZZZ}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choosen}\:\mathrm{from}\:\mathrm{8}\:\mathrm{digits} \\ $$$$\Rightarrow\mathrm{3}×\mathrm{4}×\mathrm{8}×\mathrm{7}×\mathrm{6} \\ $$$$\mathrm{case}\:\mathrm{2}: \\ $$$$\mathrm{Y}\:\mathrm{is}\:\mathrm{one}\:\mathrm{from}\:\mathrm{2}\:\mathrm{digits}\:\mathrm{in}\:\left\{\mathrm{1},\mathrm{9}\right\} \\ $$$$\mathrm{X}\:\mathrm{can}\:\mathrm{be}\:\mathrm{one}\:\mathrm{from}\:\mathrm{5}\:\mathrm{digits}\:\mathrm{in}\:\left[\mathrm{3},\mathrm{7}\right] \\ $$$$\mathrm{ZZZ}\:\mathrm{can}\:\mathrm{be}\:\mathrm{choosen}\:\mathrm{from}\:\mathrm{8}\:\mathrm{digits} \\ $$$$\Rightarrow\mathrm{2}×\mathrm{5}×\mathrm{8}×\mathrm{7}×\mathrm{6} \\ $$$$ \\ $$$$\mathrm{totally} \\ $$$$\Rightarrow\mathrm{3}×\mathrm{4}×\mathrm{8}×\mathrm{7}×\mathrm{6}+\mathrm{2}×\mathrm{5}×\mathrm{8}×\mathrm{7}×\mathrm{6}=\mathrm{7392} \\ $$
Commented by mrW1 last updated on 27/Jul/17
3×4×8×7×6+2×5×8×7×6=7392
$$\mathrm{3}×\mathrm{4}×\mathrm{8}×\mathrm{7}×\mathrm{6}+\mathrm{2}×\mathrm{5}×\mathrm{8}×\mathrm{7}×\mathrm{6}=\mathrm{7392} \\ $$
Commented by Tinkutara last updated on 27/Jul/17
Thank you very much mrW1 Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{mrW1}\:\mathrm{Sir}! \\ $$

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