Question

What shall be the arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in 1421 including itself?

• A. 3333
• B. 2448
• C. 2222
• D. 2442
• E. 2592

Hint:

For permutations with repetition, sum of all numbers = (sum of digits in each position) $\times$ (111...1).

Answer 1

The Correct Option is C

Step 1: Digits: 1, 4, 2, 1. Distinct permutations: Total permutations = $\frac{4!}{2!} = 12$ distinct numbers.
Step 2: Sum of all distinct numbers formed by digits where repetition exists: For each position, the sum of digits appearing in that position across all permutations can be found.
Step 3: Sum of all digits = 1+4+2+1 = 8. Each digit appears in each position equally often. Number of times each digit appears in a specific position = $\frac{\text{Total permutations} \times \text{frequency of digit}}{\text{total digits}} = \frac{12 \times f_d}{4}$.
Step 4: For digit 1: frequency = 2, so appearances per position = $\frac{12 \times 2}{4} = 6$. For digit 2: frequency = 1, appearances = $\frac{12 \times 1}{4} = 3$. For digit 4: frequency = 1, appearances = $\frac{12 \times 1}{4} = 3$.
Step 5: Sum in each position = $6 \times 1 + 3 \times 2 + 3 \times 4 = 6 + 6 + 12 = 24$.
Step 6: Total sum of all numbers = $24 \times (1000 + 100 + 10 + 1) = 24 \times 1111 = 26664$.
Step 7: Number of distinct numbers = 12. Mean = $\frac{26664}{12} = 2222$.
Step 8: Final Answer: The arithmetic mean is 2222.