Introduction to Counting Problems
Counting problems often appear complex, but they can usually be broken down using the Fundamental Counting Principle. This principle states that if there are $n$ ways to do one thing, and $m$ ways to do another, then there are $n \times m$ ways to do both.
The Problem
We need to find how many numbers exist between 100 and 1000 where every digit is restricted to being either a 2 or a 9.
Step-by-Step Solution
- Identify the constraints: Numbers between 100 and 1000 are all three-digit numbers. Thus, we have three positions to fill:
[Hundreds][Tens][Units]. - Analyze each position:
- For the hundreds place, we have 2 choices: {2, 9}.
- For the tens place, we have 2 choices: {2, 9}.
- For the units place, we have 2 choices: {2, 9}.
- Apply the Multiplication Principle: Since each choice is independent, we multiply the number of possibilities for each position together: $$Total = 2 \times 2 \times 2 = 2^3 = 8$$
- Verification: The numbers are: 222, 229, 292, 299, 922, 929, 992, 999. All eight are between 100 and 1000.
Conclusion
By breaking the number into independent slots, we quickly determined that there are 8 possible integers satisfy the criteria.