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

We have to form four-digit numbers with the multiset of digits {2, 2, 1, 3}.
Because two 2’s are identical, the number of distinct arrangements is $$\frac{4!}{2!}=12$$.

Let us compute the required sum by analysing the contribution of each digit in each place (thousands, hundreds, tens, ones).

Step 1: How many numbers have a given digit in a fixed place?
Fix one position, say the thousands place.
• If the digit there is 1, the remaining digits are {2, 2, 3}. Distinct permutations of these three digits $$=\frac{3!}{2!}=3$$.
• If the digit there is 3, the remaining digits are {2, 2, 1}. Again $$3$$ permutations.
• If the digit there is 2, after placing one 2 we are left with {2, 1, 3} (all distinct). Permutations of three distinct digits $$=3!=6$$.
So, in one fixed place, 1 appears 3 times, 3 appears 3 times and 2 appears 6 times.

Step 2: Contribution to the sum from each place
Thousands place (weight $$1000$$):
$$1\cdot3\cdot1000 + 3\cdot3\cdot1000 + 2\cdot6\cdot1000 = 3000 + 9000 + 12000 = 24000$$.

Because every place is symmetrical, the same frequency counts apply to the hundreds, tens and ones places.
• Hundreds place (weight $$100$$): $$1\cdot3\cdot100 + 3\cdot3\cdot100 + 2\cdot6\cdot100 = 300 + 900 + 1200 = 2400$$.
• Tens place (weight $$10$$): $$1\cdot3\cdot10 + 3\cdot3\cdot10 + 2\cdot6\cdot10 = 30 + 90 + 120 = 240$$.
• Ones place (weight $$1$$): $$1\cdot3\cdot1 + 3\cdot3\cdot1 + 2\cdot6\cdot1 = 3 + 9 + 12 = 24$$.

Step 3: Total sum
Add the contributions from all four places:
$$24000 + 2400 + 240 + 24 = 26664$$.

Hence, the sum of all distinct four-digit numbers that can be formed with the digits 2, 1, 2, 3 is $$\mathbf{26664}$$.