Any suggestion on solving this problem to get rid of double/over counting?
The digits $1$, $2$, $2$, $3$, and $4$ are placed on separate cards. How many different $3$-digit numbers can be formed by arranging the cards?
I tried... the case where all the numbers are distinct....like $1,2,3,4,5$ to make $3$-digit # and that is $5 \cdot 4 \cdot 3 = 60$ possibilities. Now I need to correct for over counting as $2$ is a repeated digit. Any help is appreciated.