The sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is equal to

To solve this problem, we need to understand that each digit will appear in each position (thousands, hundreds, tens, ones) the same number of times.

Step 1: Calculate the total number of four-digit numbers that can be formed using the digits 2, 1, 2, 3. Since there are 4 digits but one of them is repeated, we use the formula for permutations of a multiset: 4! / 2! = 12.

Step 2: Calculate the sum for each position. Each digit will appear in each position 12 / 4 = 3 times. So, the sum for each position is (2+1+2+3) * 3 = 24.

Step 3: Calculate the total sum. The sum of all the numbers is the sum for each position times the value of the position (thousands, hundreds, tens, ones). So, the total sum is 24 * (1000 + 100 + 10 + 1) = 24 * 1111 = 26664.

So, the sum of all the four-digit numbers that can be formed using all the digits 2, 1, 2, 3 is 26664.