How do you find the coefficient of $x^3y^2$ in the expansion of $(x-3y)^5$ ?

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Answer 1 · 2017-01-20T07:50:01

The coefficient of $x^3y^2$ in $(x-3y)^5$ is $90$ .

Expansion of $(a+b)^n$ gives us $(n+1)$ terms which are given by

binomial expansion $color(white)x ^nC_ra^((n-r))b^r$ , where $r$ ranges from $n$ to $0$ .

Note that powers of $a$ and $b$ add up to $n$ and in the given problem this $n=5$ .

In $(x-3y)^5$ , we need coefficient of $x^3y^2$ , we have $3^(rd)$ power of $x$ and as such $r=5-3=2$

and as such the desired coefficient of $x^3y^2$ is given by

$color(white)x ^5C_2x^((5-2))(-3y)^2=(5xx4)/(1xx2)x^3(-3y)^2$

= $10x^3xx9y^2=90x^3y^2$

Hence, the coefficient of $x^3y^2$ in $(x-3y)^5$ is $90$ .

How do you find the coefficient of x^3y^2x^3y^2 in the expansion of (x-3y)^5(x-3y)^5?