# Why is x= _3C_9 impossible to evaluate?

##### 1 Answer
Apr 2, 2018

It's not impossible to evaluate: it's just 0.

The best way to think of ${\setminus}_{n} {C}_{r}$ is as "n choose r", or "how many ways can I choose r things from n things?"

In your case, that would mean "how many ways can I choose 9 things from 3 things?" If I only have 3 things, there is no way I can choose 9 things. Hence, there are 0 possible ways to do it.

If you wanted to consider ${\setminus}_{9} {C}_{3}$, we can easily calculate that:
\ _9C_3 = (9!)/(3!6!) = (9*8*7)/(3*2*1) = 3 * 4 * 7 = 84