Permutations of "n" Distinct Items:
A Six, Two and Jack are three distinct cards. A rearrangement of the order in which these cards appear is called a "permutation". How many possible orderings are there? There are six permutations:

1. Six, Two, Jack
2. Six, jack, Two
3. Two, Six, Jack
4. Two, Jack, Six
5. Jack, Two, six
6. Jack, Six, Two

In general, the number of different permutations of "n" distinct items is:

n! = n(n-1)(n-2) . . . 1



Permutations of "r" Out of "n" Distinct Items:
Now consider a somewhat different problem. We again have n distinct items, but now we only want to use r of them.

In general, the number of different permutations of "r" out of "n" distinct items is:

n(n-1)(n-2) . . . (n - r + 1) = n!/(n-r)!