How many odd 4-digit numbers can be formed if...?

Question

How many odd 4-digit numbers, all of the digits different, can be formed using the digits 0-9 if the digit 5 must be included?

Answer 1

Say 5 is in the first place. That leaves:
4 odd possibilities for the 4th place,
8 possibilities for the 2nd place,
7 possibilities for the 3rd place
1 x 8 x 7 x 4 = 224

Say 5 is in the 2nd place. That leaves:
4 odd possibilities for the 4th place,
7 possiblilites for the 1st place, (no zero in 1st place)
7 possibilities for the 3rd place
7 x 1 x 7 x 4 = 196

Say 5 is in the 3rd place. That leaves:
4 odd possibilities for the 4th place,
7 possiblilites for the 1st place, (no zero in 1st place)
7 possibilities for the 2nd place
7 x 7 x 1 x 4 = 196

Say 5 is in the 4th place. That leaves:
8 possiblilites for the 1st place, (no zero in 1st place)
8 possibilities for the 2nd place
7 possibilities for the 3rd place
8 x 8 x 7 x 1 = 448

Total = 224 + 196 + 196 + 448 = 1,064

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If zero is allowed in the 1st position ( boo, that's not math!),
then we can increase each of the possibilites for the 1st
place (where 5 is in the 2nd, 3rd or 4th position only). In
summary we get the following:

1 x 8 x 7 x 4 = 224
8 x 1 x 7 x 4 = 224
8 x 7 x 1 x 4 = 224
9 x 8 x 7 x 1 = 504

Total = 1,176 if zero is allowed as first digit.

Answer 2

I had to write a short program to find the answers.
If the first digit can be a zero, there are 1176 such numbers.
If the first digit cannot be zero, there are 1064 such numbers.