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 [tex]
n! = \prod_{r=1}^n r[/tex] for [itex]n > 0[/itex] gives the familiar [tex]
n! = n \times (n-1) \times \dots \times 2 \times 1.[/tex] But for [itex]n = 0[/itex] it gives the empty product which by convention is [itex]\displaystyle \prod_{r \in \emptyset} r = 1[/itex]. 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.