Originally Posted by
star_tenshi Think about each digit of the PIN individually, how many numbers can there possibly be?
0, 1, 2, 3, 4, 5, 6, 7, 8, 9 = 10 possibilities
Say the first digit is a 0, move on to the second digit, this digit can, again, have 10 possibilities.
So for each possibility of the first digit, there are another 10 possibilites for the second digit. This would correspond to: 10 x 10 in terms of math.
Then there would be another ten possibilities for the third digit of the PIN, and the same for the fourth digit of the PIN.
In total, you would have: 10 x 10 x 10 x 10 = 10,000 possibilities.
This is true when you're trying to figure out the number of possibilites of anything else where repetition is allowed. If there are n number of possibilities for m spots (in this case 10 possibilities for 4 spots), then the answer would be $\displaystyle n^m$ possibilities.