how many 4 digit odd numbers can be formed with 0,3,5,4 and 2 if repetition of digits is allowed

Question

How to go about solving this problem? My calculation:
Odd numbers should end with 5 or 3.
I)With 5
remaining vacant positions are 3 which can be filled in $4.5.5=100$ ways
II)With 3
remaining vacant positions are 3 which can be filled in $4.5.5=100$ ways
Total of 200 ways.Is this right?

Answer 1

Correct, but you can simplify even further:

• You can put one of $4$ digits on the thousands place
• You can put any digit on the hundreds place
• You can put any digit on the tens place
• You can put one of $2$ digits on the ones place

So the result is $$4\cdot 5\cdot 5\cdot 2=200$$