Question

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.

Answer 1

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 `(n-i)`, and all of the numbers below `(n-i)`, 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(n-1)*...*(n-i+1)`,

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