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

1 Answer
Dec 17, 2017

Answer:

#""^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:
#(cancel(12)*11*cancel(10)*9*8)/(cancel(5*4*3*2))=11*9*8=99*8=792#