Math Problem Statement

How many 4-digit numbers have non-increasing digits?

Solution

To find how many 4-digit numbers have non-increasing digits, we need to count all possible 4-digit numbers where each digit is greater than or equal to the next digit. Let's represent a 4-digit number as abcdabcd where abcda \geq b \geq c \geq d and each of aa, bb, cc, and dd are digits from 0 to 9.

Solution Outline

  1. Define the Range for Each Digit: Since we are dealing with 4-digit numbers, aa (the thousands place) must be non-zero (i.e., a1a \geq 1), while bb, cc, and dd can each be any digit from 0 to 9.

  2. Apply Combinatorics: We need to find combinations of digits such that they are non-increasing. To do this, we can use a combinatorial approach with the concept of "stars and bars" or "combinations with repetition."

  3. Transform the Problem: Let's define a new variable to count how often each digit appears in our number. For instance, let:

    • x1x_1 represent the count of 9s,
    • x2x_2 the count of 8s,
    • x3x_3 the count of 7s,
    • ...
    • x10x_{10} represent the count of 0s.

    These counts satisfy: x1+x2++x10=4x_1 + x_2 + \cdots + x_{10} = 4 because the total number of digits in any 4-digit number is 4.

  4. Constraints: Ensure that the first non-zero digit appears in the thousands place to avoid 0 starting the number. We can handle this by subtracting combinations that start with zero.

  5. Calculate Using Combinations with Repetition: The number of solutions to the equation x1+x2++x10=4x_1 + x_2 + \cdots + x_{10} = 4 (where each xi0x_i \geq 0) is given by the formula for combinations with repetition: (4+101101)=(139)=715\binom{4 + 10 - 1}{10 - 1} = \binom{13}{9} = 715

After accounting for restrictions on the first digit (ensuring it’s non-zero), this approach provides the count of all valid non-increasing 4-digit numbers.

Final Answer

There are 715 four-digit numbers with non-increasing digits.

Would you like more details on the steps?

  1. How do we handle non-increasing sequences in other lengths, like 5-digit numbers?
  2. Can we use similar combinatorial approaches for non-decreasing digits?
  3. How do we calculate combinations with constraints on specific digits?
  4. Could we explore examples of 4-digit numbers with repeating digits?
  5. What’s the role of “stars and bars” in combinatorial math?

Tip: Non-increasing sequences follow a systematic pattern, making them ideal for combinatorics-based counting.

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Math Problem Analysis

Mathematical Concepts

Combinatorics
Permutations and Combinations
Non-increasing sequences

Formulas

Combinations with repetition formula: C(n + r - 1, r - 1)

Theorems

Stars and Bars Theorem

Suitable Grade Level

Grades 11-12