Suppose that a "code" consists of 4 digits, none of which are repeated. (a digit is one of the 10 numbers 0,1,2,3,4,5,6,7,8,9) how many codes are possible?

The answer is 5040.

This is permutation without the repetition:

P = n! / (n - r) !

n - the number of choices

r - the number of chosen

We have:

n = 10

r = 4

P = 10! / (10 - 4) !

= 10!/6!

= (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / (6 * 5 * 4 * 3 * 2 * 1)

= 10 * 9 * 8 * 7

= 5040