Counting Integers with Exactly One Digit 1 and One Digit 3 from 0 to 9999
Understanding the intricacies of digit placement in numbers is a fundamental aspect of combinatorics and number theory. This article delves into a specific problem: counting the number of integers from 0 to 9999 that have exactly one digit 1 and exactly one digit 3.
Breakdown of the Problem
We will systematically analyze the problem by considering different lengths of the numbers: 4 digits, 3 digits, 2 digits, and 1 digit. Each case will be examined in detail to ensure a comprehensive solution.
Step 1: Determine the Length of the Number
We are looking at numbers ranging from 0000 to 9999. We need to consider cases based on the number of digits in the number.
Case 1: 4-digit numbers (1000 to 9999)
The first step is to choose positions for the digits 1 and 3. There are 4 positions available, and we need to choose 2 of them to place 1 and 3. The number of ways to choose 2 positions from 4 is given by the binomial coefficient:
binom{4}{2} 6
For each selection of positions, there are 2 ways to arrange 1 and 3 (either 1 in the first chosen position and 3 in the second or vice versa). The remaining 2 positions can be filled with any digit from 0 to 9, excluding 1 and 3, which leaves us with 8 choices for each position:
text{Total for 4-digit numbers} 6 times 2 times 8^2 6 times 2 times 64 768
Case 2: 3-digit numbers (100 to 999)
For 3-digit numbers, we can choose 2 positions from 3, which gives us (binom{3}{2} 3):
Again, there are 2 ways to arrange 1 and 3.
The last position can be filled with any digit from 0 to 9, except 1 and 3, giving us 8 choices:
text{Total for 3-digit numbers} 3 times 2 times 8 48
Case 3: 2-digit numbers (10 to 99)
For 2-digit numbers, we can choose 2 positions from 2, which gives us (binom{2}{2} 1):
There are 2 ways to arrange 1 and 3.
There are no remaining positions to fill.
text{Total for 2-digit numbers} 1 times 2 2
Case 4: 1-digit numbers (0 to 9)
This case is not applicable because a single-digit number cannot have both 1 and 3.
Step 3: Summing the Cases
Now we sum the totals from each case:
textbf{4-digit numbers: 768, 3-digit numbers: 48, 2-digit numbers: 2}
Total: (text{768 48 2} 818)
Final Answer
Therefore, the total number of integers from 0 to 9999 that have exactly one digit 1 and exactly one digit 3 is (boxed{818}).
Conclusion
This step-by-step approach demonstrates the practical application of combinatorics in solving problems related to digit placement. It provides a clear and structured way to approach similar problems in future.