# Find co-efficient of $x^{5}y^{4}$ is the expansion of $\left ( \frac{1}{2}x-2y \right )^{90}$ Option 1) 126 Option 2) 63 Option 3) 84 Option 4) None of these

As.learnt in concept

General Term in the expansion of (x+a)^n -

$T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}$

- wherein

Where $r\geqslant 0 \, and \, r\leqslant n$

$r= 0,1,2,----n$

$(\frac{1}{2}x-2y)^{9}$

General term = $^9C_r$ $(\frac{1}{2}x)^{9-r}(-2y)^{r}$

Here r=4

Thus we get, $^9C_4(\frac{1}{2})^5 x^{5}*(-2)^{4}y^{4}$

$=\frac{9!}{5!4!}*\frac{1}{2^{5}}*(-2)^{4}x^{5}y^{4}$

$=63x^{5}y^{4}$

Option 1)

126

Incorrect option

Option 2)

63

Correct option

Option 3)

84

Incorrect option

Option 4)

None of these

Incorrect option

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