Challenging Counting Problem

[knight247]

If repetition of digits is not allowed, then how many 5 digit numbers begin with a digit that is prime and end with a digit that is odd?

Don't have answer options for this one. Detailed explanations would be appreciated. Thanks

[sam2304]

Begin with even prime
1 even prime. 5 odd digits to end the number and rest to fill the remaining

_ _ _ _ _
1 8 7 6 5 = 1680


Begin with odd prime - 3 5 7
3 odd prime. 4 odd digits to end the number other than the odd prime with which it is started and rest to fill the remaining

_ _ _ _ _
3 8 7 6 4 = 4032

Total = 5712 ways.

Hope i am right. Not good at P&C