The arithmetic mean of all the distinct numbers that can be obtained by rearranging the digits in $1421$, including itself, is

We need to find the average of all distinct numbers formed by rearranging the digits of 1421.

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1421 contains: 1, 4, 2, 1

The digit 1 appears twice, while 4 and 2 each appear once.

When some digits repeat, we can't just use 4! because that would count identical arrangements multiple times.

Number of arrangements = $\frac{n!}{a! \times b! \times ...}$

Where n = total number of digits and a, b, ... = number of times each digit repeats.

Number of arrangements = $\frac{4!}{2!} = \frac{24}{2} = 12$

So we have 12 distinct numbers.

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Each digit appears in each position (thousands, hundreds, tens, units) the same number of times across all arrangements.

Since digit 1 appears twice in the original number:

• Digit 1 will appear in each position: $\frac{12 \times 2}{4} = 6$ times
• Digits 4 and 2 each appear in each position: $\frac{12 \times 1}{4} = 3$ times

In thousands place: $(1 \times 6 + 4 \times 3 + 2 \times 3) \times 1000 = (6 + 12 + 6) \times 1000 = 24000$

In hundreds place: $(1 \times 6 + 4 \times 3 + 2 \times 3) \times 100 = 24 \times 100 = 2400$

In tens place: $(1 \times 6 + 4 \times 3 + 2 \times 3) \times 10 = 24 \times 10 = 240$

In units place: $(1 \times 6 + 4 \times 3 + 2 \times 3) \times 1 = 24 \times 1 = 24$

Total sum = 24000 + 2400 + 240 + 24 = 26664

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Arithmetic Mean = $\frac{\text{Sum of all numbers}}{\text{Count of numbers}} = \frac{26664}{12} = 2222$