Animated Solution for Mathematics - Permutations and Combinations: The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to ____.

We are tasked with finding the sum of all four-digit numbers formed using the set $\{1,2,2,3\}$. The total number of digits is $n=4$, but the digit $2$ repeats $p=2$ times.

If all digits were distinct, we would have $4!=24$ permutations. Due to the repetition of the digit $2$, the total number of unique four-digit numbers is:

This value represents the size of our sample space.

Instead of listing every number, we utilize the property of place value. Any four-digit number can be expressed as $1000a+100b+10c+d$, where $a,b,c,$ and $d$ represent the digits in their respective positions.

The total sum of all such numbers is the sum of the values contributed by each column. If we determine the sum of the digits in the units column, we can apply that same sum to the tens, hundreds, and thousands columns due to symmetry.

To find the sum of the digits in the units place, we calculate how many times each digit appears in that position:

1. If we fix $3$ at the units place, the remaining digits are $\{1,2,2\}$. The number of ways to arrange these is $\frac{3!}{2!}=3$. Thus, $3$ appears $3$ times.

2. If we fix $2$ at the units place, the remaining digits are $\{1,2,3\}$. Since these are distinct, the number of arrangements is $3!=6$. Thus, $2$ appears $6$ times.

3. If we fix $1$ at the units place, the remaining digits are $\{2,2,3\}$. The number of arrangements is $\frac{3!}{2!}=3$. Thus, $1$ appears $3$ times.

Now, we calculate the sum of the digits in the units column:

Because of the symmetry of the permutations, this sum is identical for the tens, hundreds, and thousands places. The total sum is therefore:

Factoring out the $24$, we obtain:

The final result is 26664.