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8-digit-numbers-are-formed-using-1-1-2-2-2-3-4-4-The-number-of-such-numbers-in-which-the-odd-digits-do-not-occupy-odd-places-is-a-160-b-120-c-60-d-48-




Question Number 52278 by Necxx last updated on 05/Jan/19
8 digit numbers are formed using  1 1 2 2 2 3 4 4. The number of such  numbers in which the odd digits  do not occupy odd places is  a)160 b)120 c)60 d)48
$$\mathrm{8}\:{digit}\:{numbers}\:{are}\:{formed}\:{using} \\ $$$$\mathrm{1}\:\mathrm{1}\:\mathrm{2}\:\mathrm{2}\:\mathrm{2}\:\mathrm{3}\:\mathrm{4}\:\mathrm{4}.\:{The}\:{number}\:{of}\:{such} \\ $$$${numbers}\:{in}\:{which}\:{the}\:{odd}\:{digits} \\ $$$${do}\:{not}\:{occupy}\:{odd}\:{places}\:{is} \\ $$$$\left.{a}\left.\right)\left.\mathrm{1}\left.\mathrm{60}\:{b}\right)\mathrm{120}\:{c}\right)\mathrm{60}\:{d}\right)\mathrm{48} \\ $$
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
$$\underset{\mathrm{8}} {−}\:\underset{\mathrm{7}} {−}\:\underset{\mathrm{6}} {−}\:\underset{\mathrm{5}\:} {−}\:\underset{\mathrm{4}} {−}\:\underset{\mathrm{3}} {−}\:\underset{\mathrm{2}} {−}\:\underset{\mathrm{1}} {−} \\ $$$${odd}\:{place}\:{are}\left(\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}\right) \\ $$$${even}\:{place}\:\left(\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right) \\ $$$${odd}\:{digit}\left(\mathrm{1},\mathrm{1},\mathrm{3}\right)\rightarrow{three} \\ $$$${even}\:{digit}\:\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right)\rightarrow{five} \\ $$$${now} \\ $$$${place}\:\mathrm{1},\mathrm{3},\mathrm{5},\mathrm{7}\:\:{to}\:{be}\:{filled}\:{by}\:{even}\:{digits}\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right) \\ $$$${so}\:{numbers}\:{of}\:{way}\:{is}=\frac{\mathrm{5}×\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{3}!×\mathrm{2}!}=\mathrm{10} \\ $$$$\left[{denominator}\:\:\mathrm{3}!\:{for}\:\left(\mathrm{2},\mathrm{2},\mathrm{2}\right)\:\:{and}\:\mathrm{2}!\:{for}\left(\mathrm{4},\mathrm{4}\right)\:\right] \\ $$$$ \\ $$$$ \\ $$$${remaning}\:{four}\:{place}\:\left(\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{8}\right) \\ $$$${rrmaining}\:{digit}\:\left(\mathrm{1},\mathrm{1},\mathrm{3}\right){nd}\:{one}\:{even}\:{digit} \\ $$$${ways}=\frac{\mathrm{4}×\mathrm{3}×\mathrm{2}×\mathrm{1}}{\mathrm{2}!}=\mathrm{12} \\ $$$${out}\:{of}\:{these}\:\left(\mathrm{10}×\mathrm{12}\right)=\mathrm{120} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 06/Jan/19
what is your final result sir?
$${what}\:{is}\:{your}\:{final}\:{result}\:{sir}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jan/19
thank you sir...clue is in your answer...
$${thank}\:{you}\:{sir}…{clue}\:{is}\:{in}\:{your}\:{answer}… \\ $$
Commented by mr W last updated on 06/Jan/19
thanks for confirming sir!
$${thanks}\:{for}\:{confirming}\:{sir}! \\ $$
Commented by Necxx last updated on 06/Jan/19
I just love this explanation.Thank  you sir Tanmay.
$${I}\:{just}\:{love}\:{this}\:{explanation}.{Thank} \\ $$$${you}\:{sir}\:{Tanmay}. \\ $$
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 1,1, 3  can only be placed in 4 even positions.    to arrange the three odd digits (1,1,3)  in 4 even positions there are  (P_3 ^4 /(2!)) different ways.  to arrange the remaining 5 even  digits (2,2,2,4,4) in the remaining 5  possitions there are  ((5!)/(2!3!)) different ways.  so the result is (P_3 ^4 /(2!))×((5!)/(2!3!))=((4×3×2)/2)×((5×4)/2)=120    ⇒answer b) is correct
$${the}\:{only}\:{restriction}:\:{odd}\:{digits}\:{don}'{t}\:{occupy} \\ $$$${odd}\:{places},\:{i}.{e}.\:{the}\:{three}\:{digits}\:\mathrm{1},\mathrm{1},\:\mathrm{3} \\ $$$${can}\:{only}\:{be}\:{placed}\:{in}\:\mathrm{4}\:{even}\:{positions}. \\ $$$$ \\ $$$${to}\:{arrange}\:{the}\:{three}\:{odd}\:{digits}\:\left(\mathrm{1},\mathrm{1},\mathrm{3}\right) \\ $$$${in}\:\mathrm{4}\:{even}\:{positions}\:{there}\:{are} \\ $$$$\frac{{P}_{\mathrm{3}} ^{\mathrm{4}} }{\mathrm{2}!}\:{different}\:{ways}. \\ $$$${to}\:{arrange}\:{the}\:{remaining}\:\mathrm{5}\:{even} \\ $$$${digits}\:\left(\mathrm{2},\mathrm{2},\mathrm{2},\mathrm{4},\mathrm{4}\right)\:{in}\:{the}\:{remaining}\:\mathrm{5} \\ $$$${possitions}\:{there}\:{are} \\ $$$$\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!}\:{different}\:{ways}. \\ $$$${so}\:{the}\:{result}\:{is}\:\frac{\mathrm{P}_{\mathrm{3}} ^{\mathrm{4}} }{\mathrm{2}!}×\frac{\mathrm{5}!}{\mathrm{2}!\mathrm{3}!}=\frac{\mathrm{4}×\mathrm{3}×\mathrm{2}}{\mathrm{2}}×\frac{\mathrm{5}×\mathrm{4}}{\mathrm{2}}=\mathrm{120} \\ $$$$ \\ $$$$\left.\Rightarrow{answer}\:{b}\right)\:{is}\:{correct} \\ $$
Commented by Necxx last updated on 06/Jan/19
yeah....very explanatory.Thanks
$${yeah}….{very}\:{explanatory}.{Thanks} \\ $$

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