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Question Number 52276 by Necxx last updated on 05/Jan/19
An8digitnumberdivisibleby9istobeformedusingdigitsfrom0to9withoutrepeatingthedigits.Thenumberofwaysinwhichthiscanbedoneis?a)72(7i)b)18(7i)c)40(7i)d)36(7i)pleaseImeanfactorialbyi.Thepartofmyscreenwherethefactorialsymbolisisntfunctioning.Thanks
Answered by mr W last updated on 06/Jan/19
wehave10digits,butweneedonly8fromthemtoforma8−digitnumber.thatistosay,wemusttake2ofthe10digitsaway.butwhichdigitsshallwetakeaway?itisrequestedthatthe8−digitnumbersaredivisibleby9.weknowanumberisdivisibleby9onlywhenthesumofitsdigitsisdivisibleby9.since0+1+2+...+9=45,itmeanswhenwetakeall10digits,thenumbersformedbythemarealwaysdivisibleby9,since45isdivisibleby9.whenwewanttoremovetwodigitstoform8−digitnumberswhichremaindivisibleby9,thesumofthesetwodigitsmustbedivisibleby9.thereareonlyfollowing5possibilities:wetake0and9away,wetake1and8away,wetake2and7away,wetake3and6away,wetake4and5away.whenwetake0and9away,withtheremaining8digits1to8wecanformtotally8!numbers.whenwetake1and8away,withtheremaining8digits(oneofthemis0)wecanformtotallyonly7×7!numberssince0cannotbeplacedinthefirstposition.sincewehave4suchsimilarwaystotaketwodigitsaway,wecantotallyform4×7×7!numbers.sotheresultis8!+4×7×7!=36×7!⇒answerd)iscorrect.
Commented by Necxx last updated on 06/Jan/19
wow.....Thankyousir.
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