# How do you find the binomial coefficient of ""^12C_5?

Dec 17, 2017

""^12C_5=792

#### Explanation:

I will use the factorial definition of ""^nC_r:
""^nC_r=((n),(r))=(n!)/(r!(n-r)!)

Plugging in the numbers we get:
""^12C_5=(12!)/(5!*7!)

To cancel with the 7! on the bottom, I'll rewrite the top:
(12*11*10*9*8*cancel(7!))/(5!*cancel(7!))

Now I'll expand the 5! and start to cancel:
$\frac{\cancel{12} \cdot 11 \cdot \cancel{10} \cdot 9 \cdot 8}{\cancel{5 \cdot 4 \cdot 3 \cdot 2}} = 11 \cdot 9 \cdot 8 = 99 \cdot 8 = 792$