# Counting phone numbers

1. Nov 12, 2004

### THA

Problem details:
I think if I pretend that N is also a number from 0-9, then there would be P(10, 7) numbers without repeats, but then that's too much and I'm not sure how I can subtract the number's that N is not supposed to have. The other way I was thinking is the last 6 numbers can be chosen P(10, 6) without repeat's but then I'm not sure how I can choose the number of N. Can someone provide some help?

2. Nov 12, 2004

### NateTG

How many single digit numbers with no repeated digits?
What about two, three, four etc.?

Also, it might be easier to figure out how many possible 7 digit numbers there are, and then subtract off the ones starting with 0 or 1.

3. Nov 12, 2004

### THA

It's regular 7-digit phone number, none of the 7 digit's can repeat. If that's not what you meant, then please clarify.

That's what I was thinking, but I'm not really sure how to count the number's that start with 0 or 1. Would it be something like: P(10,7) - 2*P(10,6) ?

4. Nov 12, 2004

### Gokul43201

Staff Emeritus
Here's what NateTG means : Forget for now that you have a 7 digit number and look at the following.

How many ways of finding a 1-digit number ? That's easy : 10 ways. Any number from 0..9 can be used.
How many 2-digit numbers without repeats ? Okay, I can put any of the10 numbers in the ten's place, and having used up this number, I can put any of the remaining 9 numbers in the unit's place. So there's 90 possibilities or P(10,2)
In this 2-digit number, if the ten's digit had to be only one of 2..9, I can fill that spot in 8 ways, and the units spot in...still only 9 ways.

And so on...