How many 12 digit numbers can be formed without repitition using the numbers 1 to 9?
If that is the case, then there are no such numbers due to the Pigeonhole Principle.
The OP more likely means how many different 12 digit numbers can be formed with the 9 digits. But then, that is trivial, since all of them can be formed with the 9 digits, all $\displaystyle 10^{13}-10^{12}$ of them i think.
I am not sure I understand where this number is coming from. After all, there is no 0 among available digits.The OP more likely means how many different 12 digit numbers can be formed with the 9 digits. But then, that is trivial, since all of them can be formed with the 9 digits, all $\displaystyle 10^{13}-10^{12}$ of them i think.
You are right! I missed that fact. Here is what I think the answer is!
You have 12 positions and 9 numbers. Each position can hold any of those 9 numbers. This gives that the number of such numbers is
$\displaystyle 9\times9\times9\times9\ldots\times9 = 9^{12} = 282,429,536,481$
by the multiplication principle of combinatorics. Now, if you restrict yourself to using the digits without repetition, you run into a problem. For the first number you can use any of the 9 numbers. Then once you choose one, you move to the next digit, but now you cannot use the first digit, so you have only 8 options for it. If you go on, the 9th digit will exhaust all the available numbers. Therefor you cannot form any 12 digit numbers with only 9 digits without repetition(so there are only 0 such numbers).
Then maybe your question is different.