# How do you evaluate 9C3?

Oct 27, 2017

$84$

#### Explanation:

in general

color(white)(x)^nC_r=(n!) /(r!(n-r)!

color(white)(x)^9C_3=(9!) /(3!(9-3)!

=(9!)/(3!6!)

=(9xx8xx7xxcancel(6!))/ (3!cancel(6!)

$= \frac{{\cancel{9}}^{3} \times {\cancel{8}}^{4} \times 7}{\cancel{3} \times \cancel{2} \times 1}$

$= 3 \times 4 \times 7 = 84$

Oct 27, 2017

$84$

See steps below;

#### Explanation:

"Formular" rArr ^nC_r = (n!)/((n - r)! r!)

Where.... $c = 9 , r = 3$

rArr ^9C_3 = (9!)/((9 - 3)!3!)

color(white)(xxxx) rArr (9!)/((9 - 3)!3!)

color(white)(xxxx) rArr (9!)/((6)!3!)

color(white)(xxxx) rArr (9!)/(6!3!)

$\textcolor{w h i t e}{\times \times} \Rightarrow \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{\left(6 \times 5 \times 4 \times 3 \times 2 \times 1\right) \left(3 \times 2 \times 1\right)}$

$\textcolor{w h i t e}{\times \times} \Rightarrow \frac{9 \times 8 \times 7 \times \cancel{6} \times \cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}}{\left(\cancel{6} \times \cancel{5} \times \cancel{4} \times \cancel{3} \times \cancel{2} \times \cancel{1}\right) \left(3 \times 2 \times 1\right)}$

$\textcolor{w h i t e}{\times \times} \Rightarrow \frac{9 \times 8 \times 7}{3 \times 2 \times 1}$

$\textcolor{w h i t e}{\times \times} \Rightarrow \frac{504}{6}$

$\textcolor{w h i t e}{\times \times} \Rightarrow 84$