The sum of all the four-digit numbers that ca | vedclass

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(C) The given digits are $1, 2, 2, 3$. The total number of four-digit numbers that can be formed is $\frac{4!}{2!} = \frac{24}{2} = 12$.To find the su

Vedclass · Accepted Answer

$26664$

Vedclass · Answer

(C) The given digits are $1, 2, 2, 3$. The total number of four-digit numbers that can be formed is $\frac{4!}{2!} = \frac{24}{2} = 12$.To find the sum of these numbers,we calculate the sum of the digits at each place (units,tens,hundreds,thousands).For any specific position,the frequency of each digit is:- Digit $1$: $\frac{3!}{2!} = 3$ times.- Digit $2$: $\frac{3!}{1!1!} = 6$ times.- Digit $3$: $\frac{3!}{2!} = 3$ times.Sum of digits at any place $= (1 	imes 3) + (2 	imes 6) + (3 	imes 3) = 3 + 12 + 9 = 24$.The sum of all such numbers is $24 	imes 1000 + 24 	imes 100 + 24 	imes 10 + 24 	imes 1 = 24 	imes 1111 = 26664$.

The sum of all the four-digit numbers that can be formed using all the digits $2, 1, 2, 3$ is equal to $.......$.

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