tag:blogger.com,1999:blog-2787161678592839240.post1102774225302827096..comments2024-03-19T12:46:33.223+05:30Comments on www.mathsblog.in : Maths Blog for School Teachers & Students: എട്ടാം ക്ലാസ് ഗണിതശാസ്ത്ര അദ്ധ്യാപകസഹായിയിലെ ഒരു സംശയംവി.കെ. നിസാര്http://www.blogger.com/profile/14303804236214732024noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-2787161678592839240.post-64767641651519813062009-08-14T10:19:36.657+05:302009-08-14T10:19:36.657+05:30ഇവിടെ ഈ ഫാക്ടോറിയലിനെന്താണ് പ്രസക്തി?
രാജ്കുമാര്...ഇവിടെ ഈ ഫാക്ടോറിയലിനെന്താണ് പ്രസക്തി?<br /><br />രാജ്കുമാര് <br />വളവുകോട്Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-73541270064308446172009-08-13T23:36:54.725+05:302009-08-13T23:36:54.725+05:30ncr=ncn-r,
put r=n
ncn=nc0
1=ni/0i*(n-0)i
it follo...ncr=ncn-r,<br />put r=n<br />ncn=nc0<br />1=ni/0i*(n-0)i<br />it follows that 0i=1<br /><br /><br />Muraleedharan. C.R<br />G.V.H.S.S VattenadAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-2787161678592839240.post-61776243151452388702009-08-13T20:08:15.704+05:302009-08-13T20:08:15.704+05:30If n! is defined as the product of all positive in...If n! is defined as the product of all positive integers from 1 to n, then:<br />1! = 1*1 = 1<br />2! = 1*2 = 2<br />3! = 1*2*3 = 6<br />4! = 1*2*3*4 = 24<br />...<br />n! = 1*2*3*...*(n-2)*(n-1)*n<br />and so on.<br />Logically, n! can also be expressed n*(n-1)! .<br /><br />Therefore, at n=1, using n! = n*(n-1)!<br />1! = 1*0!<br />which simplifies to 1 = 0!<br /> <br />This solution is from google Search<br /> <br />Regards<br /> AjeeshAnonymousnoreply@blogger.com