Combination With Repetition4 Comments gtm on December 26, 2012 at 7:30 pm Can you pls elaborate this ? = Coefficient of x14 in the expansion of (x1+ x2+ x3+ ………..+ x6)3= Coefficient of x14 in the expansion of x3(1+x1+ x2+ ………..+ x5)3= Coefficient of x14-3 in the expansion of (1+x1+ x2+ ………..+ x5)3= Coefficient of x11 in the expansion of(As 1, x, x2….x5 are in G.P.)= Coefficient of x11 in the expansion of (1 -3 )(1+ 3x + 6×2+ ………..+21 x5+……78×11+…..)=78-21(3)=15 Reply neena on June 1, 2012 at 10:34 am is there a particular method of understanding clearly how we can evaluate the coefficient of x^14 in the expansion of( x^o+ x^1 …..+x^6)^3 – the first problem .? Reply Amit Chhabra on May 10, 2012 at 7:40 am Undoubtedly Praiseworthy..What an ideas SIR JEE !But i think some problem in solution 2 :As maximum marks given = 10 , SO the all beggars could have maximum 10 coins (solution says 20)Plus we will find the coeff. of x^20 instead of x^14 .Please check and do the needful.Expecting more articles more frequently .Thanks . Reply prashant on May 11, 2012 at 2:45 am ahh thanks alot amit will look into the error u pointed out and get back to you asap…! Submit a Comment Cancel replyYour email address will not be published. Required fields are marked *Comment *Name * Email * Website
gtm on December 26, 2012 at 7:30 pm Can you pls elaborate this ? = Coefficient of x14 in the expansion of (x1+ x2+ x3+ ………..+ x6)3= Coefficient of x14 in the expansion of x3(1+x1+ x2+ ………..+ x5)3= Coefficient of x14-3 in the expansion of (1+x1+ x2+ ………..+ x5)3= Coefficient of x11 in the expansion of(As 1, x, x2….x5 are in G.P.)= Coefficient of x11 in the expansion of (1 -3 )(1+ 3x + 6×2+ ………..+21 x5+……78×11+…..)=78-21(3)=15 Reply
neena on June 1, 2012 at 10:34 am is there a particular method of understanding clearly how we can evaluate the coefficient of x^14 in the expansion of( x^o+ x^1 …..+x^6)^3 – the first problem .? Reply
Amit Chhabra on May 10, 2012 at 7:40 am Undoubtedly Praiseworthy..What an ideas SIR JEE !But i think some problem in solution 2 :As maximum marks given = 10 , SO the all beggars could have maximum 10 coins (solution says 20)Plus we will find the coeff. of x^20 instead of x^14 .Please check and do the needful.Expecting more articles more frequently .Thanks . Reply
prashant on May 11, 2012 at 2:45 am ahh thanks alot amit will look into the error u pointed out and get back to you asap…!
Can you pls elaborate this ?
= Coefficient of x14 in the expansion of (x1+ x2+ x3+ ………..+ x6)3
= Coefficient of x14 in the expansion of x3(1+x1+ x2+ ………..+ x5)3
= Coefficient of x14-3 in the expansion of (1+x1+ x2+ ………..+ x5)3
= Coefficient of x11 in the expansion of
(As 1, x, x2….x5 are in G.P.)
= Coefficient of x11 in the expansion of (1 -3 )(1+ 3x + 6×2+ ………..+21 x5+……78×11+…..)
=78-21(3)
=15
is there a particular method of understanding clearly how we can evaluate the coefficient of x^14 in the expansion of( x^o+ x^1 …..+x^6)^3 – the first problem .?
Undoubtedly Praiseworthy..What an ideas SIR JEE !
But i think some problem in solution 2 :
As maximum marks given = 10 , SO the all beggars could have maximum 10 coins (solution says 20)
Plus we will find the coeff. of x^20 instead of x^14 .
Please check and do the needful.
Expecting more articles more frequently .
Thanks .
ahh thanks alot amit
will look into the error u pointed out and get back to you asap…!