# I don't know how this happen!!! Help me in fatorial?

## I would like to understand how this happens... i!((n),(i)) = n(n-1)...(n-i+1) I know that i!((n),(i)) = i! frac{n!}{i!(n-i)!}, but I don't know how to get this equality above.

Apr 21, 2018

Cancelling the relevant factorials gives us the equality. See below.

#### Explanation:

Let's look at (i!)(n!)/((i!)(n-i)!).

The i! cancel out:

cancel((i)!)(n!)/(cancel((i!))(n-i)!)=(n!)/((n-i)!)

Furthermore, recall that n! = n(n-1)(n-2)(n-3)*...*3*2*1

Well, somewhere before n! "finishes" and reaches multiplication by $1$, we multiply by $\left(n - i\right)$, and all of the numbers below $\left(n - i\right)$, which is denoted by (n-i)!:

n! = n(n-1)(n-2)(n-3)*...*(n-i)!

Then, (n-i)! must also cancel out. This would leave us with

(n!)/((n-i)!)=(n(n-1)(n-2)(n-3)*...*cancel((n-i)!))/cancel((n-i)!)

So, we're left with

$n \left(n - 1\right) \cdot \ldots \cdot \left(n - i + 1\right)$,

because since (n-i)! cancelled out, we only multiply down to the quantity that comes right before (n-i)!, which is $\left(n - i + 1\right) .$