Counting 4-digit Odd Numbers from a Given Set

When faced with the challenge of forming 4-digit odd numbers using the digits 0, 1, 2, 3, 4, 5 without repeating any digit, we must carefully consider the constraints and utilize combinatorial mathematics to arrive at the solution. This article will guide you through the step-by-step process and present a detailed analysis.

Understanding the Problem

For a number to be odd, its last digit must be 1, 3, or 5. We start by identifying these possible last digits and then determine the available choices for the first digit and the middle two digits based on these constraints. This requires a systematic approach and the application of concepts from combinatorics.

Step-by-Step Solution

We will approach this problem by breaking it down into manageable steps:

Identify Possible Last Digits: Since the number must be odd, the last digit can be either 1, 3, or 5. There are 3 choices for the last digit.

Determine Choices for the First Digit: The first digit cannot be 0 and cannot be the same as the last digit. This leaves us with 4 available digits for the first digit, depending on the choice of the last digit. Specifically:

If the last digit is 1, the first digit can be any of 2, 3, 4, 5, giving us 4 choices. If the last digit is 3, the first digit can be any of 1, 2, 4, 5, giving us 4 choices. If the last digit is 5, the first digit can be any of 1, 2, 3, 4, giving us 4 choices.

Choose the Remaining Digits: After selecting the first and last digits, we have 2 digits left to choose from. These digits can be arranged in the two middle positions in 4P2 (permutations of 4 items taken 2 at a time) ways, which equals 12 arrangements.

Each combination of first and last digits results in 12 different arrangements for the middle digits. Therefore, the total number of 4-digit odd numbers that can be formed is calculated as follows:

[3 times 4 times 12 144]

Verifying with Python Code

To verify our solution, we can write a Python script to generate all possible 4-digit numbers and count the ones that meet the criteria.

from itertools import permutations# Define the digitsdigits  [0, 1, 2, 3, 4, 5]# Generate all 4-digit permutationsperms  permutations(digits, 4)count  0for perm in perms:    # Check if the first digit is not 0 and the last digit is odd    if perm[0] ! 0 and perm[3] in [1, 3, 5]:        count   1print(count)

This code should output the result: 144, confirming our manual calculation.

Conclusion

By following these steps and applying combinatorial principles, we have determined that there are 144 possible 4-digit odd numbers that can be formed from the given set of digits 0, 1, 2, 3, 4, 5 without repeating any digits. This method can be generalized for similar problems involving permutations and digit constraints.