Question

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

Answer 1

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

Explanation:

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`.