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Question Number 124709 by benjo_mathlover last updated on 05/Dec/20

Between 20000 and 70000 find the number of even integers in which no digits is repeated

$${Between}\:\mathrm{20000}\:{and}\:\mathrm{70000}\: \\ $$$${find}\:{the}\:{number}\:{of}\:{even}\:{integers} \\ $$$${in}\:{which}\:{no}\:{digits}\:{is}\:{repeated} \\ $$

Answered by liberty last updated on 05/Dec/20

Let abcde be a required even integer. As shown in the following diagram, the 1^(st) digit can be chosen from { 2,3,4,5,6} and the 5^(th) digit e can be chosen from {0,2,4,6,8}. ⌊ a ⌋ ⌊ b c d ⌋ ⌊e ⌋ since {2,3,4,5,6} ∩ { 0,2,4,6,8}= {2,4,6} we devide the problem into 2 disjoint cases case(1) a ∈ {2,4,6} there are = 3×4×P _3^8 case(2) a ∈ {3,5} there are = 2×5×P _3^8 The total number of required even numbers is 4032 + 3360 = 7392 .

$${Let}\:{abcde}\:{be}\:{a}\:{required}\:{even}\:{integer}.\: \\ $$$${As}\:{shown}\:{in}\:{the}\:{following}\:{diagram},\:{the} \\ $$$$\mathrm{1}^{{st}} \:{digit}\:{can}\:{be}\:{chosen}\:{from}\:\left\{\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\} \\ $$$${and}\:{the}\:\mathrm{5}^{{th}} \:{digit}\:{e}\:{can}\:{be}\:{chosen}\:{from}\: \\ $$$$\left\{\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right\}. \\ $$$$\:\lfloor\:\:{a}\:\rfloor\:\lfloor\:{b}\:{c}\:{d}\:\rfloor\:\lfloor{e}\:\rfloor\: \\ $$$${since}\:\left\{\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6}\right\}\:\cap\:\left\{\:\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right\}=\:\left\{\mathrm{2},\mathrm{4},\mathrm{6}\right\} \\ $$$${we}\:{devide}\:{the}\:{problem}\:{into}\:\mathrm{2}\:{disjoint}\:{cases} \\ $$$${case}\left(\mathrm{1}\right)\:{a}\:\in\:\left\{\mathrm{2},\mathrm{4},\mathrm{6}\right\}\:{there}\:{are}\:=\:\mathrm{3}×\mathrm{4}×{P}\:_{\mathrm{3}} ^{\mathrm{8}} \\ $$$${case}\left(\mathrm{2}\right)\:{a}\:\in\:\left\{\mathrm{3},\mathrm{5}\right\}\:{there}\:{are}\:=\:\mathrm{2}×\mathrm{5}×{P}\:_{\mathrm{3}} ^{\mathrm{8}} \\ $$$${The}\:{total}\:{number}\:{of}\:{required}\:{even}\:{numbers} \\ $$$${is}\:\mathrm{4032}\:+\:\mathrm{3360}\:=\:\mathrm{7392}\:. \\ $$

Answered by mr W last updated on 06/Dec/20

ABCDE A: 2,3,4,5,6 E: 0,2,4,6,8 case 1: E∈{0,8} ⇒2×5×P_3 ^8 =3360 case 2: E∈{2,4,6} ⇒3×4×P_3 ^8 =4032 ⇒totally 3360+4032=7392

$${ABCDE} \\ $$$${A}:\:\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{5},\mathrm{6} \\ $$$${E}:\:\mathrm{0},\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8} \\ $$$${case}\:\mathrm{1}:\:{E}\in\left\{\mathrm{0},\mathrm{8}\right\} \\ $$$$\Rightarrow\mathrm{2}×\mathrm{5}×{P}_{\mathrm{3}} ^{\mathrm{8}} =\mathrm{3360} \\ $$$${case}\:\mathrm{2}:\:{E}\in\left\{\mathrm{2},\mathrm{4},\mathrm{6}\right\} \\ $$$$\Rightarrow\mathrm{3}×\mathrm{4}×{P}_{\mathrm{3}} ^{\mathrm{8}} =\mathrm{4032} \\ $$$$ \\ $$$$\Rightarrow{totally}\:\mathrm{3360}+\mathrm{4032}=\mathrm{7392} \\ $$

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