Counting Numbers with At Most Three Different Digits from 1 to 9

Have you ever wondered how many numbers with at most three different digits can be formed from the digits 1 to 9? This seemingly simple question can lead to a detailed exploration of combinatorics and permutations. Let#8217;s dive into the solution and understand the nuances involved.

Initial Calculations

First, let's start with the basic calculations. For each digit in the number, there are 9 possible choices (1 to 9). Consequently:

One-digit numbers: 9 possibilities (1, 2, 3, 4, 5, 6, 7, 8, 9) Two-digit numbers: Since each position can be any of the 9 digits, there are 9 x 9 81 possibilities Three-digit numbers: Each of the three positions can be any of the 9 digits, resulting in 9 x 9 x 9 729 possibilities

The total count of numbers with at most three different digits is the sum of these possibilities:

1-digit: 9, 2-digit: 81, 3-digit: 729, Total: 9 81 729 819

Interpreting “Ascending”

The term "ascending" can be ambiguous. Depending on how it is interpreted, the number of valid combinations can change. For instance, some interpretations require strictly increasing digits, like 123, 234, 345, etc.

Strictly Ascending Digits

If "ascending" means strictly increasing without any repetition, the numbers 123, 234, 345, 456, 567, 678, and 789 are valid. Therefore, there are 7 such numbers.

General Case

For a more general case, we need to account for different scenarios:

None of the digits are the same: The number of ways to choose 3 distinct digits from 9 is given by the combination formula C(9, 3) Exactly two of the digits are the same and one is different: The number of ways to choose the digit that appears twice (9 ways) and the different digit (8 ways), leading to 9 x 8 ways All three digits are the same: There are 9 possible choices (111, 222, etc.)

The total number of permutations is the sum of these cases:

None the same: C(9, 3) 84

Exactly two the same: 9 x 8 72

All three the same: 9

Total: 84 72 9 165

Verification with Scala Code

To validate our calculations, we can use a simple Scala function. The following functions can be used to check if a number is valid:

scala def noZeros(num: Int): Boolean !("0") scala def digitsAscending(num: Int): Boolean

Using these functions, we can filter and count the numbers that satisfy both conditions:

val res (100 to 999).filter(noZeros).filter(digitsAscending).size val res4: Int 165

Therefore, the number of valid numbers is 165.

Conclusion

Understanding the nuances of the question and using combinatorial methods or programming can provide accurate results. This exercise not only demonstrates the beauty of mathematics but also highlights the importance of precise language in problem statements.

For further inquiries, you can explore similar problems involving combinations and permutations, which are fundamental topics in discrete mathematics and computer science. Happy coding and problem-solving!