Counting Three-Digit Numbers with One Odd and Two Even Digits: A Comprehensive Guide

Understanding how to count three-digit numbers that consist of one odd digit and two even digits is a fascinating exercise in combinatorics. This article will explore the various methods to calculate such numbers and provide a clear, step-by-step explanation. By the end of this guide, you will be well-equipped to handle similar counting problems with ease.

Introduction to the Problem

Three-digit numbers can be represented as XYZ, where X is the most significant digit (hundreds place), Y is the middle digit (tens place), and Z is the least significant digit (units place). We are interested in counting the numbers where one digit is odd and the other two are even. The even digits are {0, 2, 4, 6, 8} and the odd digits are {1, 3, 5, 7, 9}. Repetition of digits is allowed.

Method 1: Permutations and Exclusions

The first method to solve this problem involves calculating all possible permutations and then excluding those that do not meet the criteria.

Odd Digit Placement: There are 5 choices for the odd digit (1, 3, 5, 7, 9). Even Digit Placement: There are 5 choices for each of the even digits (0, 2, 4, 6, 8).

The total number of three-digit numbers formed by this method is (5 times 5 times 5 125).

Exclusions: We need to exclude the cases where all three digits are the same. These cases are 111, 333, 555, 777, 999. There are 5 such cases.

Final Calculation: The total number of three-digit numbers consisting of one odd digit and two even digits is (125 - 5 180).

Method 2: Detailed Breakdown of Permutations

We can break down the permutations into three scenarios:

Two Even Digits and One Odd Digit: If the even digits are first and first, the odd digit is third: (4C1 times 5C1 times 5C1 4 times 5 times 5 100) If the even digits are first and second, the odd digit is second: (5C1 times 5C1 times 4C1 5 times 5 times 4 125) If the even digits are second and first, the odd digit is third: (4C1 times 5C1 times 5C1 4 times 5 times 5 100) Total Calculation: The total number of three-digit integers is (100 125 100 325).

Additional Methods

There are three forms in which we can arrange a 3-digit number with 2 even digits and 1 odd digit:

EVEN EVEN ODD EVEN ODD EVEN ODD EVEN EVEN

In the first two forms, we have 4 choices (2, 4, 6, 8) for the first digit (which cannot be 0), and 5 choices for the other two digits. In the third form, we have 5 choices (1, 3, 5, 7, 9) for the first digit, and 5 choices for the other two digits. The total number of such 3-digit numbers is (24 times 5 times 5 15 times 5 times 5 15 times 5 times 5 200 125 125 325).

Conclusion

With these methods, we can accurately determine that there are 325 three-digit numbers consisting of one odd digit and two even digits. This comprehensive guide provides a clear understanding of the problem and the methods to solve it effectively. Whether you are a student or a professional, this guide will help you tackle similar problems with confidence.

Further Reading

To deepen your understanding of combinatorics and number theory, consider exploring related topics such as permutations, combinations, and advanced counting techniques. By applying these concepts, you can solve a wide range of mathematical and real-world problems.