If the first digit cannot be a 0 or a 5, how many five-digit odd numbers are there?

A. 42,500
B. 37,500
C. 45,000
D. 40,000
E. 50,000

Soln: This problem can be solved with the Multiplication Principle. The Multiplication Principle tells us that the number of ways independent events can occur together can be determined by multiplying together the number of possible outcomes for each event.

There are 8 possibilities for the first digit (1, 2, 3, 4, 6, 7, 8, 9).
There are 10 possibilities for the second digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the third digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 10 possibilities for the fourth digit (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)
There are 5 possibilities for the fifth digit (1, 3, 5, 7, 9)

Using the Multiplication Principle:

= 8 * 10 * 10 * 10 * 5
= 40,000
(D).


I understand the solution and problem. However, I guess I don't completely understand it because I used 8!10!10!10!5! instead of the Multiplication Principle stated above.

My reasoning was that there are x! number of ways to arrange each possible digit.

Please guide! Thank you very much.