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 May 20th, 2011, 05:17 AM #1 Member   Joined: May 2011 Posts: 51 Thanks: 0 Binomial expansion as combination Hi all, I m new here In one book the author says to look at the binomial theorem as the algebraic codification of a combinatorial process. In (x+y)^n=(x+y)*(x+y)*(x+y)...n times Consider them as 'n' numbers of factors, consider 'r' as a positive integer between 0 and n. Now any r numbers of factors among those n factors will give us x^r and the rest will give us y^n-r, of the term x^r*y^n-r. Now those r factors can be chosen in nCr ways. So the co-efficient of x^r*y^n-r is nCr. I know this is right but I don't understand how... How come the co-efficient is coming as nCr? I don't understand how the co-efficient and the combination of r among n factors are linked? Cheers R
 May 20th, 2011, 08:38 AM #2 Senior Member     Joined: Feb 2010 Posts: 706 Thanks: 141 Re: Binomial expansion as combination Look for example at $(x+y)^4= (x+y)(x+y)(x+y)(x+y)$. Expanded it equals $x^4+4x^3y+6x^2y^2+4xy^3+y^4$ Now look at just the term $4x^3y$ and think about where the $y$ came from. It could have come from the first parenthesis ... that is $y \cdot x \cdot x \cdot x$. It could have come from the second parenthesis ... that is $x \cdot y \cdot x \cdot x$. It could have come from the third parenthesis ... that is $x \cdot x \cdot y \cdot x$. It could have come from the fourth parenthesis ... that is $x \cdot x \cdot x \cdot y$. So the coefficient 4 is really just counting the possible arrangements of x's and y's in $x^3y$. That is, the coefficient is $_4C_3$.
 May 20th, 2011, 04:22 PM #3 Member   Joined: May 2011 Posts: 51 Thanks: 0 Re: Binomial expansion as combination Thanks a lot, I got it Cheers R

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