# Permutation proof

1. Oct 24, 2015

### Pual Black

1. The problem statement, all variables and given/known data
Hello
I have to proof this
$(n+1)nPr=(n+1)P(r+1)$

The attempt at a solution
if i substitute this in
$nPr=\frac{n!}{(n-r)!}$
I get this
$(n+1)nPr=\frac{[(n+1)n]!}{[(n+1)n-r]!}$
This will give
$(n+1)nPr=\frac{(n^2+n)!}{(n^2+n-r)!}$

Now i don't know how to continue.

2. Oct 24, 2015

### SteamKing

Staff Emeritus
I don't think (n + 1) * n! = [n * (n + 1)]!

What is (n + 1)*n! ?

3. Oct 24, 2015

### haruspex

Further to SteamKing's point, why did you also attempt to multiply the denominator by n+1?

4. Oct 25, 2015

### Pual Black

I just substituted (n+1)n in the equation of nPr.
$nPr=\frac{n!}{n-r}$
Just changed the (n) with (n+1)n
I didn't multipy the denominator.

But now i think i get you
Did you mean that (n+1)nPr isn't the same as ((n+1)n)Pr ??

5. Oct 25, 2015

### haruspex

Yes. It's (n+1)(nPr).

6. Oct 25, 2015

### Pual Black

Ok thanks. Now lets go on
$(n+1)nPr=(n+1)\frac{n!}{(n-r)!}$

I have to get this
$\frac{(n+1)!}{[(n+1)-(r+1)]!}$
to proof the R.H.S

But if i start with the Right side
$(n+1)P(r+1)=\frac{(n+1)!}{[(n+1)-(r+1)]!}=(n+1)\frac{n!}{(n-r)!}=(n+1)nPr=L.H.S$
But how to start with left side and proof the right side ?

I can -1 +1 to the denominator to get the (n+1)-(r+1)
But can i say that (n+1)n!=(n+1)! ?
I think yes of course

7. Oct 25, 2015

### haruspex

Yes. And since you are dealing with equalities, not inequalities, it doesn't matter which side you start with to arrive at the other.

8. Oct 25, 2015

### Pual Black

Ok. Thank you very much for your help.