In combinatorics, we often need to calculate the sum and average of all possible N-digit numbers formed by arranging N distinct digits exactly once. For example, with digits 1, 2, 3, we can form 3-digit numbers like 123, 132, 213, 231, 312, and 321. This article explores the formula for the sum, demonstrates manual counting, and provides step-by-step solutions to practice problems.
The formula for the sum of all possible N-digit numbers formed by N distinct digits is:
The average is the sum divided by the total number of such numbers, which is \( N! \).
For example, the sum of all numbers formed with digits 1, 2, 3, 4 is \( 10 \times 3! \times 1111 = 66,660 \).
Manual counting is practical for small N but becomes tedious for larger N. Let’s compute the sum for N=2 with digits 1 and 2.
Calculate the sum of all 2-digit numbers formed using digits 1, 2 exactly once.
Solution:
Step 1: List all possible numbers.
Step 2: Compute the sum.
\( 12 + 21 = 33 \)
Step 3: Verify with the formula.
Sum = \( 3 \times 1 \times 11 = 33 \)
The formula matches the manual sum!
Answer: The sum is 33.
For larger N, such as N=3 (digits 1, 2, 3), listing all \( 3! = 6 \) numbers (123, 132, 213, 231, 312, 321) and summing them manually is time-consuming. The formula is far more efficient.
Question 1: What is the average of all five-digit numbers formed using the digits 1, 2, 3, 4, 5 exactly once?
Solution:
To find the average, we calculate the sum of all possible numbers and divide by the total number of numbers.
Step 1: Determine the total number of numbers.
Number of 5-digit numbers = \( 5! = 120 \).
Step 2: Calculate the sum of digits.
Sum of digits = \( 1 + 2 + 3 + 4 + 5 = 15 \).
Step 3: Apply the sum formula: \( (\text{Sum of digits}) \times (N-1)! \times (111\ldots\text{N times}) \)
For \( N = 5 \):
Sum = \( 24 \times 15 \times 11111 \)
\( 24 \times 15 = 360 \)
\( 360 \times 11111 = 3,999,960 \)
Step 4: Calculate the average.
Average = \( \frac{3,999,960}{120} = 33,333 \)
Answer: The average is 33,333.
Question 2: Find the sum of all four-digit numbers formed using the digits 2, 4, 6, 8 exactly once.
Solution:
Step 1: Determine the total number of numbers.
Number of 4-digit numbers = \( 4! = 24 \).
Step 2: Calculate the sum of digits.
Sum of digits = \( 2 + 4 + 6 + 8 = 20 \).
Step 3: Apply the sum formula.
For \( N = 4 \):
Sum = \( 20 \times 6 \times 1111 \)
\( 20 \times 6 = 120 \)
\( 120 \times 1111 = 133,320 \)
Answer: The sum is 133,320.
Question 3: What is the average of all three-digit numbers formed using the digits 1, 3, 5 exactly once?
Solution:
Step 1: Determine the total number of numbers.
Number of 3-digit numbers = \( 3! = 6 \).
Step 2: Calculate the sum of digits.
Sum of digits = \( 1 + 3 + 5 = 9 \).
Step 3: Apply the sum formula.
For \( N = 3 \):
Sum = \( 9 \times 2 \times 111 \)
\( 9 \times 2 = 18 \)
\( 18 \times 111 = 1,998 \)
Step 4: Calculate the average.
Average = \( \frac{1,998}{6} = 333 \)
Answer: The average is 333.
Question 4: Find the sum of all six-digit numbers formed using the digits 1, 2, 3, 4, 5, 6 exactly once.
Solution:
Step 1: Determine the total number of numbers.
Number of 6-digit numbers = \( 6! = 720 \).
Step 2: Calculate the sum of digits.
Sum of digits = \( 1 + 2 + 3 + 4 + 5 + 6 = 21 \).
Step 3: Apply the sum formula.
For \( N = 6 \):
Sum = \( 21 \times 120 \times 111111 \)
\( 21 \times 120 = 2,520 \)
\( 2,520 \times 111111 = 279,999,720 \)
Answer: The sum is 279,999,720.
Calculating the sum and average of N-digit numbers is a powerful application of combinatorics. The formula \( (\text{Sum of digits}) \times (N-1)! \times (111\ldots\text{N times}) \) simplifies the process, especially for large N. Practice with the examples provided to master this concept, and explore its applications in exams, programming, and beyond.