Consider
$$A = \{ 1,2,3,4,5,6\}$$
a) How many $3$ digit even numbers between $200$ and $500$ can be formed?
b) How many $3$ digit odd numbers between $240$ and $600$ can be formed?
For the part a, we have that
$$P = (5)(6)(3) = 90$$
For the part b, we have that
$$P = (5)(2)(3) = 30$$
Am I wrong?
EDIT: Repetation is ALLOWED.
Regards.
For $(a)$, they have first digit $2,3$ or $4$, and last digit $2,4$ or $6$. So, $3\cdot 6\cdot 3=54$.
For $(b)$, first digit can be $2,3,4$ or $5$. Second digit can be $4,5$ or $6$ when the first digit is $2$, and the last digit is $1,3$ or $5$. So $3\cdot3=9$ for when the first digit is two. Then when the first digit is $3,4$ or $5$, we have any of the six for the second, and $1,3$ or $5$ for the third. So, $3\cdot 6\cdot 3=54$. So we get $54+9=63$.