How Many 3-Digit Numbers Have Their Digits in Increasing Order?

Understanding the distribution of 3-digit numbers with digits in increasing order is a fascinating combinatorial problem. This article explores how many such numbers exist and provides the mathematical reasoning behind it. We will also delve into a brute force method using the J programming language and include a simple C program for demonstration.

Introduction

When dealing with the set of digits {1, 2, 3, 4, 5, 6, 7, 8, 9}, how many 3-digit numbers can we form such that the digits are in increasing order? Let's explore the solution step-by-step.

Step-by-Step Solution

Choosing 3 Distinct Digits

First, we need to choose 3 distinct digits from the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Note that the digit 0 is excluded as it cannot be the leading digit in a 3-digit number. Thus, we are left with 9 digits to choose from.

Arranging the Digits

Once we have chosen our 3 digits, we need to arrange them in increasing order. Since there is only one way to do this for any set of 3 different digits, we don't need to consider permutations.

Calculation

The number of ways to choose 3 digits from 9 can be calculated using the combination formula:

[ binom{9}{3} frac{9!}{3!(9-3)!} ]

Simplifying this gives:

[ binom{9}{3} frac{9 times 8 times 7}{3 times 2 times 1} frac{504}{6} 84 ]

Therefore, there are 84 3-digit numbers whose digits are in increasing order.

Brute Force Solution Using J Programming Language

The J programming language provides a straightforward way to solve this problem using a combinatorial approach. Here is an example of a J script that identifies all such numbers:

code
  /m._1 _1-/
  3
  /The answer is that there are 7 even three-digit positive integers that have strictly increasing digits left to right.
  /List them:
  mn
  234 456 678
/code

This script first defines a verb `m` to check if the digits of a number are in strictly increasing order. It then filters the list of 3-digit numbers to find those that satisfy this condition. The result shows that there are 3 even numbers among these 84 numbers.

Brute Force Solution Using C Programming Language

To further illustrate the brute force approach, here is a simple C program that can be used to identify these numbers:

code
#include stdio.h
int main() {
  int i, t  0;
  char s[5];
  for (i  100; i  1000; i  ) {
    sprintf(s, %d, i);
    if (s[0]  s[1]  s[1]  s[2]) {
      t  ;
    }
  }
  printf(There are %d 3-digit numbers with digits in increasing order., t);
  return 0;
}
/code

This program iterates through all 3-digit numbers, converts them to strings, and checks if the digits are in increasing order. It counts and prints the total number of such numbers.

Conclusion

In conclusion, there are 84 unique 3-digit numbers with digits in increasing order. This problem not only demonstrates the power of combinatorial methods but also highlights the elegance and simplicity of programming solutions in languages like J and C.