A Permutation is an ordered list of (all) unique elements of a set.

Let's say the set S is:

S:=\{0,1,2\}

Then which ways are there to lay out these elements into a new ordered list, using all of them exactly once?

0,...
1,...
2,...

and for each of these except one (the one with the same element twice):

?,0,...
?,1,...
?,2,...

and for each of these except for a few (the one with the same element twice):

?,?,0
?,?,1
?,?,2

So in total:

0,1,2
0,2,1
1,0,2
1,2,0
2,0,1
2,1,0

whoah! So only 6 remain (I would have expected a lot more).

What happened is that after you took out one element of the set to place it at the first position, the set contained one element less out out which you could draw for the next position.

If you are only interested in the number of permutations possible, it seems there should be an easier way...

And there is:

So let's have a time-dependent set S(t) which we define by:

|S(0)|:=3
|S(1)|:=2
|S(2)|:=1

Then the number of combinations is: (number of elements we could use at t=0) times (number of elements we could use at t=1) times (number of elements we could use at t=2).

So a long story short: The (number of permutations) c is:

c=3⋅2⋅1
c=|S(0)|⋅|S(1)|⋅...

If you always build it one-by-one (which you can reduce it to), then:

c=|S(0)|⋅(|S(0)|-1)⋅... (as long as it's greater than 0)

And this is called the factorial, written "!".

c=|S(0)|! Number of Permutations.