Permutations combinatorics problem?

[SMA_01]

Homework Statement

A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations

P(n,r)=n!/(n-r)!

The Attempt at a Solution

For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?

 

[Dick]

Homework Statement

A telephone extension has four digits, how many different extensions are there with no repeated digits, if the first digit cannot be zero?

Homework Equations

P(n,r)=n!/(n-r)!

The Attempt at a Solution

For the first digit, there are 9 possibilities (because no zero)
For the last 3 digits I used P(9,3)=9!/6!=9x8x7

So, in the end my result was: 9x9x8x7= 4,536 different extensions...

I'm just wondering if I was correct?

Seems fine to me.