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Question Number 52278 by Necxx last updated on 05/Jan/19
$$\mathrm{8}\:{digit}\:{numbers}\:{are}\:{formed}\:{using} \\ $$$$\mathrm{1}\:\mathrm{1}\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}.\:{The}\:{number}\:{of}\:{such} \\ $$$${numbers}\:{in}\:{which}\:{the}\:{odd}\:{digits} \\ $$$${do}\:{not}\:{occupy}\:{odd}\:{places}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\mathrm{60}\:{b}\right)\mathrm{120}\:{c}\right)\mathrm{60}\:{d}\right)\mathrm{48} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19
![−_8 −_7 −_6 −_(5 ) −_4 −_3 −_2 −_1 odd place are(1,3,5,7) even place (2,4,6,8) odd digit(1,1,3)→three even digit (2,2,2,4,4)→five now place 1,3,5,7 to be filled by even digits(2,2,2,4,4) so numbers of way is=((5×4×3×2)/(3!×2!))=10 [denominator 3! for (2,2,2) and 2! for(4,4) ] remaning four place (2,4,6,8) rrmaining digit (1,1,3)nd one even digit ways=((4×3×2×1)/(2!))=12 out of these (10×12)=120](Q52292.png)
$$\underset{\mathrm{8}} {−}\:\underset{\mathrm{7}} {−}\:\underset{\mathrm{6}} {−}\:\underset{\mathrm{5}\:} {−}\:\underset{\mathrm{4}} {−}\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{1}} {−} \\ $$$${odd}\:{place}\:{are}\left(\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}\right) \\ $$$${even}\:{place}\:\left(\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right) \\ $$$${odd}\:{digit}\left(\mathrm{1},\mathrm{1},\mathrm{3}\right)\rightarrow{three} \\ $$$${even}\:{digit}\:\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right)\rightarrow{five} \\ $$$${now} \\ $$$${place}\:\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}\:\:{to}\:{be}\:{filled}\:{by}\:{even}\:{digits}\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right) \\ $$$${so}\:{numbers}\:{of}\:{way}\:{is}=\frac{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{3}!×\mathrm{2}!}=\mathrm{10} \\ $$$$\left[{denominator}\:\:\mathrm{3}!\:{for}\:\left(\mathrm{2},\mathrm{2},\mathrm{2}\right)\:\:{and}\:\mathrm{2}!\:{for}\left(\mathrm{4},\mathrm{4}\right)\:\right] \\ $$$$ \\ $$$$ \\ $$$${remaning}\:{four}\:{place}\:\left(\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right) \\ $$$${rrmaining}\:{digit}\:\left(\mathrm{1},\mathrm{1},\mathrm{3}\right){nd}\:{one}\:{even}\:{digit} \\ $$$${ways}=\frac{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{2}!}=\mathrm{12} \\ $$$${out}\:{of}\:{these}\:\left(\mathrm{10}×\mathrm{12}\right)=\mathrm{120} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 06/Jan/19
$${what}\:{is}\:{your}\:{final}\:{result}\:{sir}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19
$${thank}\:{you}\:{sir}...{clue}\:{is}\:{in}\:{your}\:{answer}... \\ $$
Commented by mr W last updated on 06/Jan/19
$${thanks}\:{for}\:{confirming}\:{sir}! \\ $$
Commented by Necxx last updated on 06/Jan/19
$${I}\:{just}\:{love}\:{this}\:{explanation}.{Thank} \\ $$$${you}\:{sir}\:{Tanmay}. \\ $$
Answered by mr W last updated on 06/Jan/19
$${the}\:{only}\:{restriction}:\:{odd}\:{digits}\:{don}'{t}\:{occupy} \\ $$$${odd}\:{places},\:{i}.{e}.\:{the}\:{three}\:{digits}\:\mathrm{1},\mathrm{1},\:\mathrm{3} \\ $$$${can}\:{only}\:{be}\:{placed}\:{in}\:\mathrm{4}\:{even}\:{positions}. \\ $$$$ \\ $$$${to}\:{arrange}\:{the}\:{three}\:{odd}\:{digits}\:\left(\mathrm{1},\mathrm{1},\mathrm{3}\right) \\ $$$${in}\:\mathrm{4}\:{even}\:{positions}\:{there}\:{are} \\ $$$$\frac{{P}_{\mathrm{3}} ^{\mathrm{4}} }{\mathrm{2}!}\:{different}\:{ways}. \\ $$$${to}\:{arrange}\:{the}\:{remaining}\:\mathrm{5}\:{even} \\ $$$${digits}\:\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right)\:{in}\:{the}\:{remaining}\:\mathrm{5} \\ $$$${possitions}\:{there}\:{are} \\ $$$$\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!}\:{different}\:{ways}. \\ $$$${so}\:{the}\:{result}\:{is}\:\frac{\mathrm{P}_{\mathrm{3}} ^{\mathrm{4}} }{\mathrm{2}!}×\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!}=\frac{\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{2}}×\frac{\mathrm{5}×\mathrm{4}}{\mathrm{2}}=\mathrm{120} \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{b}\right)\:{is}\:{correct} \\ $$
Commented by Necxx last updated on 06/Jan/19
$${yeah}....{very}\:{explanatory}.{Thanks} \\ $$