Question

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

Answer 1

`""^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`