tag:blogger.com,1999:blog-6519442308665124636.post343412344691556735..comments2018-04-07T13:15:14.522+05:30Comments on Maths & Physics - Solved Questions: gauravragtahhttp://www.blogger.com/profile/03828604566438196779noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6519442308665124636.post-20915879003302862582009-01-14T14:56:00.000+05:302009-01-14T14:56:00.000+05:30Good work Ashwin, that's how I confirmed my answer...Good work Ashwin, that's how I confirmed my answer later. <BR/>However, on this site I post the solutions I gave on GoIIT as is; and there I just did whatever struck me then and there... the 1 term made it easy 'cos of x^0 all along.spideyunlimitedhttps://www.blogger.com/profile/03828604566438196779noreply@blogger.comtag:blogger.com,1999:blog-6519442308665124636.post-89306351390952539052009-01-14T02:49:00.000+05:302009-01-14T02:49:00.000+05:30Hmm. A quicker method would be the multinomial the...Hmm. A quicker method would be the multinomial theorem, bro.<BR/><BR/>Find the coefficient of x^4 in (1 + 2x + 3x^2)^5 .<BR/>> p + q + r = 5<BR/><BR/>> [5!/[p!*q!*r!]*(1^p)*(2^q)*(x^q)*(3^r)*(x^2r)<BR/><BR/>> Basically the powers of the x - terms have to be equal to 4.<BR/>q + 2r = 4<BR/>By elementary:<BR/>We have 3 cases, <BR/>(4,0) p = 1<BR/>(2,1) p = 2<BR/>(0,2) p = 3<BR/><BR/>Apply it in the above expression:<BR/>80 + 360 + 90 = 530Ashwin Ravichandrannoreply@blogger.com