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

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

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Answer 1 · 2017-01-20T07:50:01

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

Explanation:

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

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

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

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

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

$x^{5} C_{2} x^{(5 - 2)} {(- 3 y)}^{2} = \frac{5 \times 4}{1 \times 2} x^{3} {(- 3 y)}^{2}$ $color(white)x ^5C_2x^((5-2))(-3y)^2=(5xx4)/(1xx2)x^3(-3y)^2$

= $10 x^{3} \times 9 y^{2} = 90 x^{3} y^{2}$ $10x^3xx9y^2=90x^3y^2$

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

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

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