An-8-digit-number-divisible-by-9-is-to-be-formed-using-digits-from-0-to-9-without-repeating-the-digits-The-number-of-ways-in-which-this-can-be-done-is-a-72-7i-b-18-7i-c-40-7i-d-36-7i-please-I-m Tinku Tara June 4, 2023 Permutation and Combination 0 Comments FacebookTweetPin 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. Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Question-183344Next Next post: Find-a-lim-x-3x-2-x-2-x-1-b-lim-x-x-2-1-2x-1-c-lim-x-5-3x-1-4-x-5- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.