Question

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?

Answer 1

nah thats not correct. ^

As it is a 4 digit code, the first digit cannot be 0.. so there are 9 choices..
For the second one.. as the digits cannot be repeated, excluding the digit you used for the first number, you now have 9 choices..
For the third number similarly, you have 8 choices.
For the fourth number you have 7 choices..

So your answer would be 9*9*8*7 ..

Answer 2

The question does not include a restriction that the first digit cannot be zero. Therefore the number of permutations of the 10 digits taken 4 at a time without repetitions is given by
\[\frac{10!}{(10-4)!}\]