Counting n digit Numbers with all unique digits in JavaScript

We need to write a JavaScript function that counts all n-digit numbers where every digit appears exactly once (all digits are unique).

Problem Statement

Given a number num, find how many numbers exist with exactly num digits where all digits are unique.

For example:

• 1-digit numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ? Count = 10
• 2-digit numbers: 10, 12, 13, ..., 98 (excluding 11, 22, 33, etc.) ? Count = 91

Understanding the Logic

This is a combinatorics problem:

• 1-digit: All 10 digits (0-9) are valid
• 2-digit: First digit: 9 choices (1-9, not 0), Second digit: 9 choices (remaining digits including 0)
• 3-digit: First digit: 9 choices, Second: 9 choices, Third: 8 choices

Solution Using Dynamic Programming

const uniqueDigits = (num = 1) => {
    // Base cases: dp[i] = count of i-digit numbers with unique digits
    const dp = [1, 10];
    
    // For n >= 2, apply the formula
    for (let i = 2; i <= num; i++) {
        // First digit: 9 choices (1-9)
        // Remaining digits: decreasing choices (9, 8, 7, ...)
        dp[i] = 9;
        let availableDigits = 9;
        
        for (let j = 1; j < i; j++) {
            dp[i] *= availableDigits;
            availableDigits--;
        }
    }
    
    return dp[num];
};

// Test cases
console.log("1-digit numbers:", uniqueDigits(1));
console.log("2-digit numbers:", uniqueDigits(2));
console.log("3-digit numbers:", uniqueDigits(3));
console.log("4-digit numbers:", uniqueDigits(4));

1-digit numbers: 10
2-digit numbers: 81
3-digit numbers: 648
4-digit numbers: 4536

Alternative Mathematical Approach

const uniqueDigitsMath = (num) => {
    if (num === 0) return 1;
    if (num === 1) return 10;
    if (num > 10) return 0; // Can't have more than 10 unique digits
    
    // For n-digit numbers: 9 × 9 × 8 × 7 × ... × (11-n)
    let result = 9; // First digit choices (1-9)
    
    for (let i = 2; i <= num; i++) {
        result *= (11 - i); // Remaining digit choices
    }
    
    return result;
};

// Comparison
console.log("Mathematical approach:");
console.log("1-digit:", uniqueDigitsMath(1));
console.log("2-digit:", uniqueDigitsMath(2));
console.log("3-digit:", uniqueDigitsMath(3));
console.log("Edge case - 11 digits:", uniqueDigitsMath(11));

Mathematical approach:
1-digit: 10
2-digit: 81
3-digit: 648
Edge case - 11 digits: 0

Step-by-Step Breakdown

Key Points

• Maximum possible digits with unique digits is 10 (using all digits 0-9)
• For 1-digit numbers, we include 0 as a valid number
• For multi-digit numbers, the first digit cannot be 0
• Each subsequent position has one fewer choice than the previous

Conclusion

This problem uses combinatorics principles where we calculate permutations with constraints. The mathematical approach is more efficient than dynamic programming for this specific problem since it directly computes the result using the formula 9 × 9! / (10-n)!.