If all possible 4-digit numbers are formed by | vedclass

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(C) The number of ways to choose $4$ different digits from the set $\{1, 2, 3, 5, 8\}$ is $^5P_4 = 5 	imes 4 	imes 3 	imes 2 = 120$ numbers.Each di

Vedclass · Accepted Answer

$506616$

Vedclass · Answer

(C) The number of ways to choose $4$ different digits from the set $\{1, 2, 3, 5, 8\}$ is $^5P_4 = 5 	imes 4 	imes 3 	imes 2 = 120$ numbers.Each digit appears in each place (units,tens,hundreds,thousands) an equal number of times.Since there are $5$ digits and we choose $4$,each digit appears in a specific position $^4P_3 = 4 	imes 3 	imes 2 = 24$ times.The sum of the digits is $S = 1 + 2 + 3 + 5 + 8 = 19$.The sum of the digits in any one position is $24 	imes 19 = 456$.The sum of all such numbers is $456 	imes (1 + 10 + 100 + 1000) = 456 	imes 1111 = 506616$.

If all possible $4$-digit numbers are formed by choosing $4$ different digits from the given digits $1, 2, 3, 5, 8$,then the sum of all such $4$-digit numbers is:

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