Question

How do you evaluate 8C3?

Answer 1

`""_8C_3=56`.

Explanation:

We know that, `""_nC_r=(n!)/{(n-r)!*r!}`

With, `n=8, r=3, ((n-r)!)=5!`

So, `""_8C_3=(8!)/(5!*3!)=(cancel(5!)*6*7*8)/(cancel5!*3!)`

`=(6*7*8)/(1*2*3)=56`.