Final answer:
To form distinct integers between 300 and 3000 using digits 1, 2, and 3 only, we can create 12 unique combinations, considering the permutations of the digits in each place value.
Step-by-step explanation:
The question aims to figure out how many distinct integers between 300 and 3000 can be constructed using the digits 1, 2, and 3 only. First, we need to consider numbers with these three digits. Since the numbers should be between 300 and 3000, our smallest digit, 1, cannot be in the hundreds place. So we must have either 2 or 3 in the hundreds place to satisfy the condition that numbers must be greater than 300.
We have two options for the hundreds place (2 or 3), three options for the tens place (1, 2, or 3, whichever two are not in the hundreds place), and two options for the ones place (the remaining digit). Thus, there are 2 x 3 x 2 = 12 possible unique numbers that can be made using the digits 1, 2, and 3 when forming numbers between 300 and 3000.
This question requires an understanding of permutations, which is a fundamental concept of combinatorics in Mathematics. Additionally, it shows that mathematics often offers multiple methods to arrive at a solution, reinforcing the robustness of mathematical principles.