Question

A four digit number is made using the numbers 1 through 9. The number must start and end with an odd number, and repetition of digits is allowed. How many four-digit numbers can be made?

Answer 1

http://openstudy.com/updates/4fdbf71ae4b0f2662fd22770
I found a similar question with this response but still not sure I am understanding.
kropot72 3 years ago
b. There are 5P2 permutations of the 5 odd numbers taken 2 at a time. And there are 5P2 permutations of the even numbers taken 2 at a time (taking 0 as an even number)
Therefore there are 5P2 * 5P2 configurations:
5!3!×5!3!=5×4×5×4=400

Answer 2

Would this mean I qould have 5!4!x4!4!=5x4x4x5=400? Since it would be 5 odd x 4 even x 4 even x 5 odd?

Answer 3

I'm gonna write this up in case you come back later.

I'm not sure what you mean by using "!" factorial, but your answer of 5 * 4 * 4 * 5 is close. However, the problem says nothing about the middle digits being even. Just that the first and last be odd.

Think of this as having 4 slots to fill. And we can reuse numbers ...

___ ___ ___ ___
In the first slot, you can have a 1, 3, 5, 7, or 9 ... so 5 choices
In the 2nd, you can choose any number 1 to 9 .. so 9 choices
In the 3rd, you can choose any number 1 to 9 .. so 9 choices
In the last slot, you can have a 1, 3, 5, 7, or 9 ... so 5 choices

By multiplication principle: 5 * 9 * 9 * 5 = 2025 possible numbers.