1. ## Four-Digit Odd Numbers

How can I solve this problem without using the fundamental counting principle?

How many 4-digit odd numbers greater than 3,000 can be formed by using the digits 0-9 if digits may be repeated?

I was told this is a permutation NOT a combination word problem.

In that case, can I apply nPn or nPr. If not, why not?

2. Hello, sharkman!

How many 4-digit odd numbers greater than 3,000 can be formed
by using the digits 0-9 if digits may be repeated?
We have a four-digit number.

The first (leftmost) digit can be any of {3, 4, 5, 6, 7, 8, 9} . . . 7 choices.

The second digit can be any of the 10 digits.

The third digit can be any of the 10 digits.

The last (rightmost) digit can be any of {1, 3, 5, 7, 9} . . . 5 choices.

Therefore . . .

3. ## Then...

Originally Posted by Soroban
Hello, sharkman!

We have a four-digit number.

The first (leftmost) digit can be any of {3, 4, 5, 6, 7, 8, 9} . . . 7 choices.

The second digit can be any of the 10 digits.

The third digit can be any of the 10 digits.

The last (rightmost) digit can be any of {1, 3, 5, 7, 9} . . . 5 choices.

Therefore . . .
7*10^2*5 = 3500

Is there a way to solve this using a formula?