# How do you find the 9th term in the expansion of the binomial (10x-3y)^12?

Feb 16, 2017

${T}_{9} = 32476950000 {x}^{4} {y}^{8.}$

#### Explanation:

In the Expansion of Binomial ${\left(a + b\right)}^{n} ,$ the General ${\left(r + 1\right)}^{t h}$ Term

is given by, ${T}_{r + 1} = \text{_nC_ra^(n-r)b^r" where, } r = 0 , 1 , 2 , \ldots n .$

In our Problem,

$a = 10 x , b = - 3 y , n = 12 , \mathmr{and} , r + 1 = 9 , i . e . , r = 8.$

:. T_9=""_12C_8(10x)^(12-8)(-3y)^8

=""_12C_4(10)^4(3)^8x^4y^8,

$= \frac{\left(12\right) \left(11\right) \left(10\right) \left(9\right)}{\left(1\right) \left(2\right) \left(3\right) \left(4\right)} \left(10000\right) \left(6561\right) {x}^{4} {y}^{8}$

$= \left(11\right) \left(5\right) \left(9\right) \left(10000\right) \left(6561\right) {x}^{4} {y}^{8}$

${T}_{9} = 32476950000 {x}^{4} {y}^{8}$

Enjoy Maths.!