Question about permutation formula -- How to use the formula when r = 0 ?

[songoku]

Homework Statement

This is not homework

The formula of permutation is $P^n_r=\frac{n!}{(n-r)!}=n(n-1)...(n-r+1)$

I want to ask how to use the formula when r = 0

Relevant Equations

$P^n_r=\frac{n!}{(n-r)!}=n(n-1)...(n-r+1)$

$P^5_0=\frac{5!}{(5-0)!}=1$

But when I use $P^n_r=n(n-1)...(n-r+1)$, I get $P^5_0=5(4)...(6)$

Where is the mistake? Thanks

 

[anuttarasammyak]

You use the forlula which is applicable for r $\ge$ 1 and not applicable for r=0. 5>4>3>...>6 is obviously unreasonable.

 

[songoku]

You use the forlula which is applicable for r $\ge$ 1 and not applicable for r=0. 5>4>3>...>6 is obviously unreasonable.

Sorry I still don't really understand.

$P^n_r=n(n-1)...(n-r+1)$ is derived from $P^n_r=\frac{n!}{(n-r)!}$ and for the later formula there is no restriction $r \geq 1$

So it means we add a new restriction when deriving the formula $P^n_r=n(n-1)...(n-r+1)$?

Thanks

 

[pasmith]

The formula <br /> n! = \prod_{r=1}^n r for n &gt; 0 gives the familiar <br /> n! = n \times (n-1) \times \dots \times 2 \times 1. But for n = 0 it gives the empty product which by convention is \displaystyle \prod_{r \in \emptyset} r = 1. So it should not surprise you that formulae derived for cases where arguments of ! are strictly positive may not apply when an argument of ! is zero.

 

[songoku]

Thank you very much for the explanation anuttarasammyak and pasmith

 

[Mark44]

Another explanation is that $P^5_0=\frac{5!}{(5-0)!} = \frac{5!}{5!} = 1$
In words, the number of permutations of 5 things taken 0 at a time is, per the above, 1.

 

[haruspex]

$n(n-1)...(n-r+1)$

… means: start with 1, then multiply it by n, then by n-1, etc., but don't go below n-r+1.
With r=0, n is already below n-r+1, so stop at 1.