The sum of all 4-digit numbers that can be fo | vedclass

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(C) The given digits are $2, 3, 4, 5, 6$. There are $n = 5$ digits in total. We need to form $4$-digit numbers without repetition. The number of such

Vedclass · Accepted Answer

$533280$

Vedclass · Answer

(C) The given digits are $2, 3, 4, 5, 6$. There are $n = 5$ digits in total. We need to form $4$-digit numbers without repetition. The number of such $4$-digit numbers is $P(5, 4) = \frac{5!}{(5-4)!} = 120$. In each place (thousands,hundreds,tens,units),each digit appears an equal number of times. The number of times each digit appears in a specific place is $\frac{120}{5} = 24$ times. The sum of the digits is $S = 2 + 3 + 4 + 5 + 6 = 20$. The sum of the numbers is given by: Sum $= 24 	imes S 	imes (10^3 + 10^2 + 10^1 + 10^0)$ Sum $= 24 	imes 20 	imes (1111)$ Sum $= 480 	imes 1111 = 533280$.

The sum of all $4$-digit numbers that can be formed using the digits $2, 3, 4, 5, 6$ without repetition is:

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