How many difference 3-digit even numbers can be formed from the digits 2,3,5,7 and 9, if each digit can be used more than once?

Question

anonymous · Answer

first choice you have 4, second you have 4, third you have only one choice (a 2) so the answer is 16

anonymous · Answer

the answer would be $$5^{3}$$=125 combinations if  each one of them can  repeat

anonymous · Answer

First number you have 5 choices, second number you have 5 choices, but for the last you only have one (the 2 to make it even).  So 5*5*1 = 25.

anonymous · Answer

Lol so which one is it

anonymous · Answer

what are the 5 choices?

anonymous · Answer

the 5 digits you can pick from.
First digit can be a 2, 3, 5, 7, or 9 (5 choices)
Second digit can be a 2, 3, 5, 7, 0 (5 choices)
last digit can only be a 2 (one choice)

anonymous · Answer

thank you