The sum of all the 4-digit distinct numbers t | vedclass

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(A) The given digits are $1, 2, 2, 3$. The total number of distinct $4$-digit numbers is $\frac{4!}{2!} = 12$.To find the sum,we calculate the frequen

Vedclass · Accepted Answer

$26664$

Vedclass · Answer

(A) The given digits are $1, 2, 2, 3$. The total number of distinct $4$-digit numbers is $\frac{4!}{2!} = 12$.To find the sum,we calculate the frequency of each digit at each place (units,tens,hundreds,thousands).For any position,the number of times a digit appears is:- Digit $1$: $\frac{3!}{2!} = 3$ times.- Digit $2$: $\frac{3!}{1!} = 6$ times.- Digit $3$: $\frac{3!}{2!} = 3$ times.Sum of digits at any position $= (1 	imes 3) + (2 	imes 6) + (3 	imes 3) = 3 + 12 + 9 = 24$.Since there are $4$ positions,the total sum is $24 	imes (1000 + 100 + 10 + 1) = 24 	imes 1111 = 26664$.

The sum of all the $4$-digit distinct numbers that can be formed with the digits $1, 2, 2,$ and $3$ is:

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