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.
On
Imagine you have $\color{red}{2}$ and $\color{blue}{2}$, so that 2s are distinct. Every solution with a 2 in it has a corresponding solution where red and blue are swapped. These pairs of solutions should be considered the same; this will get rid of double-counting.
So, to avoid double-counting, you need to add up (1) the number of solutions without any twos, plus (2) half the number of solutions containing twos.
There are $3! = 6$ solutions that don't contain twos, and there are $60-6=54$ solutions containing twos. Hence the correct number of solutions, avoiding double-counting, is:
$$6 + \frac{1}{2}(54) = 6 + 27 = 33$$
Consider cases:
Three different numbers are used: There are four ways to select the hundreds digit, three ways to select the tens digit, and two ways to select the units digit. Hence, there are $4 \cdot 3 \cdot 2 = 24$ such numbers.
Two different numbers are used: There must be a repeated $2$. Choose two of the three positions for the two $2$s. Choose one of the other three digits for the free position. There are $\binom{3}{2}\binom{3}{1} = 9$ such numbers.
Total: Since the two cases above are mutually exclusive and exhaustive, there are $24 + 9 = 33$ three-digit numbers that can be formed with the given digits.