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

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To find the arithmetic mean, we calculate the sum of all distinct permutations of the digits in 1421 and divide it by the total number of permutations:

1. **Total Permutations:**  
   - The digits in 1421 are: 1, 4, 2, and another 1 (repeated).  
   - Total permutations = $$ \frac{4!}{2!} = \frac{24}{2} = 12 $$.

2. **Distribution of Digits in Each Place:**  
   - Each digit appears equally in all positions (units, tens, hundreds, thousands).  
   - For example, in the unit place:  
     - Digit '1' appears $$6$$ times.  
     - Digit '2' appears $$3$$ times.  
     - Digit '4' appears $$3$$ times.

- Total sum of digits in any single place (e.g., units):  
     $$
     	ext{Sum} = (6 	imes 1) + (3 	imes 2) + (3 	imes 4) = 24.
     $$

3. **Sum Across All Places:**  
   - The sum of digits is identical for units, tens, hundreds, and thousands places.  
   - Total sum across all places = $$24(1 + 10 + 100 + 1000) = 24(1111)$$.

4. **Arithmetic Mean Calculation:**  
   - Total sum of all permutations = $$24 	imes 1111 = 26664$$.  
   - Arithmetic mean = $$ \frac{26664}{12} = 2222 $$.

Thus, the arithmetic mean of all distinct numbers is **2222**, and the correct answer is **B**.

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