[soop]

Before we start, I don't have an answer to this, it's just something of a puzzle I thought might be interesting.

My Friend hit me up on facebook saying she'd lost her phone, and did I have her parents number, as she was coming back home. She was actually asking if I had it so I could call her, but I thought that she needed the number herself. I racked my brain, and couldn't quite get it.

So using google maps, I found the road, then looked it up in the phonebook. Not there, but the nearby numbers jogged my memory, and I realised it started with [area code] 39, then 4 extra digits. So that's 9999 possible numbers right?

Well, not quite. I know what it isn't, and that's:

The same digit 4 times in a row (e.g. 390000)

Two digits repeated (e.g. 392323, or 392233)

consecutive numbers (e.g. 391234, or 394321)

3 consecutive numbers (e.g. 391237, or 397321)

and I'm sure that the number in its entirety contains at least 4 different digits (including the 39, not the area code)

Given the above, can anyone work out the actual number of permutations it could possibly be?

And, while you could possibly sit and work it out bit by bit, I'd be interested if someone can do it mathmatically.

[Guest]

Before we start, I don't have an answer to this, it's just something of a puzzle I thought might be interesting.

My Friend hit me up on facebook saying she'd lost her phone, and did I have her parents number, as she was coming back home. She was actually asking if I had it so I could call her, but I thought that she needed the number herself. I racked my brain, and couldn't quite get it.

So using google maps, I found the road, then looked it up in the phonebook. Not there, but the nearby numbers jogged my memory, and I realised it started with [area code] 39, then 4 extra digits. So that's 9999 possible numbers right?

Well, not quite. I know what it isn't, and that's:

The same digit 4 times in a row (e.g. 390000)

Two digits repeated (e.g. 392323, or 392233)

consecutive numbers (e.g. 391234, or 394321)

3 consecutive numbers (e.g. 391237, or 397321)

and I'm sure that the number in its entirety contains at least 4 different digits (including the 39, not the area code)

Given the above, can anyone work out the actual number of permutations it could possibly be?

And, while you could possibly sit and work it out bit by bit, I'd be interested if someone can do it mathmatically.

Since 3 and 9 shown, we only left 8 numbers to be placed in position number 3. and then 7 numbers in position 4, and so on.

The Permutation should be 8 * 7 * 6 * 5 = 1,680 ?

[Prime]

...

The same digit 4 times in a row (e.g. 390000)

Two digits repeated (e.g. 392323, or 392233)

consecutive numbers (e.g. 391234, or 394321)

3 consecutive numbers (e.g. 391237, or 397321)

and I'm sure that the number in its entirety contains at least 4 different digits (including the 39, not the area code)

...

4 didgits form 10000 combinations.

The way I understand, the conditions boil down to:

No more than 3 duplicates of a digit, AND if there are 3 duplicates all 3 must not appear in a row, AND no more than 1 digit duplicated, AND no more than two neighboring digits may appear in ascending/descending sequence. It is not entiely clear whether the "9" from 39 is included in the ascending/descending prohibition, and whether "39" is a part of a double digit duplicate rule (e.g. 390139).

You are better off calling directory service.

4 ascending/descending digits is an individual case of 3-digit sequences. No need to consider separately.

There are 8 combinations for ascending sequence (from 012 to 789) and as many descending. Each 3-digit sequence has a choice of two positions to strat from. With most there are 10 different different digits possible in the remaining 4th position. Example: 39-234x.

Thus 8*2*2*10 = 320.

However, some of the combinations here we counted twice. For example: 2345 was accounted as 234x and x345.So we must subtract 16 from the number of combinatins. However, not all 3-digit strings allow double counting. I.e., 789x, x012, 210x, and x987.

And so we exclude 320 - 16 + 4 = 308 numbers out of 10000.

I did not address "39" participation in the ascending/descending rule. But you get the idea.

Digit duplicate exclusions have similar reaoning patterns and are in comparabe order.

[soop]

The 9 is included in the ascending/descending rule, and the 39 is included in the double digits rule.

Don't worry about the actual phone number, I'm not that bothered - she has my number! But still, that's some impressive work prime.