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The sum of all 4-digit numbers that can be created by arranging the digits 1, 2, 4, and 5 only once in each arrangement is

Option: 1

43292


Option: 2

32492


Option: 3

56792

 


Option: 4

79992


Answers (1)

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The number of numbers of 3 digits, having 4 in unit's place = The number of arrangement of the remaining 3 digits =  ^{3}P_3   = 6.

Similarly, for other digits also, each digit occurs 6 times in units place.

So the sum of digits in the unit's place is given by,

\begin{aligned} & S=6 \times 1+6 \times 2+6 \times 4+6 \times 5 \\ & S=6+12+24+30 \\ & S=72 \end{aligned}

Similarly, the number of digits in each of the other places = 72

Thus, the required sum is,

S=72\times 1000+72\times10+72\times1\\

S=72000+7200+720+72S=79992

Therefore, the required sum is 79992.

Posted by

shivangi.bhatnagar

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