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Question 63377: find the coefficent of x^5y^5 in the expansion of (2x+3y)^10
thank you
Answer by fomunearantess@yahoo.com(2)   (Show Source):
You can put this solution on YOUR website! we begin by stating the Binomial theorem
(a+b)^n=[S](nCr(a^n-r)b^r
for this case
a=2x,b=3y,n=10
ie
(2x+3y)^10=[S]10Cr{((2x)^(10-r)(3y)^r}=10Cr2^(10-r)x^(10-r)*3^r*y^r
[S] is used here to represent "summation of"
for coefficient of x^5y^5 we have
x^(10-r)y^r=x^5y^5
ie
10-r=5
=>r=5
using this value of r in the original expression we have the coefficient to be
10C5*2^5*3^5
ie 10!*2^5*3^5/(5!*5!)
i have left answer in factorial and index form
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