Oh and it is describing the on/off switch, what I have done with lots of on/off switches together, I can perform all the permutations of this math problem, which has been solved below using the ancient BC Chinese/Pascal's Triangle in a new way.
You have a combination padlock with four dials on it. Each dial has the numbers 0 through 4 on them. The lock can have as many 0s as dials, and is set to 0000 by default. The lock does not allow you to use any number between 1 and 4 two or more times in the combination. The following combinations are valid: 0123 1234 0103 0010 4031. The following combinations are invalid: 0113 4014 0202 4444. How many possible combinations are there?
Here is the calculus solution:
If scaled to 18 wheels and 18 numbers by changing n (which is 4 above), I can perform the 2,968,971,264,021,448,999 possible permutations each reachable within 18 keystrokes or less.
This can also be solved using Pascal's Triangle that goes back to BC ancient China in a new way, because we didn't have a representation for zero until the 13th century and in the implementation I was counting nothing as something, which is represented by zero in the word problem. The implementation is selecting 4 different grenades' order when buying and throwing, where you essentially place the grenades into the order you want to throw, with empty slots being possible if you do not select all 4 grenades.
Pascal's Triangle Solution
I have also built a randomizer that drives a relational database and calculators that can emulate addition/subtraction/multiplication/division.