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Hint: The first thing we must find out is the total number of possible arrangements of the 4 digits of the given 5, i.e., $ ^{5}{{P}_{4}} $ . Out of these arrangements, in one fifth of the cases 5 will be present in unit place, in other one fifth cases in the tens place, in another one fifth cases in the hundreds place, one fifth in the thousands place and not present in left one fifth cases. So, we can say that each term will appear at a certain place for one fifth of the total case. So,
the sum of the unit digit = $ ^{5}{{P}_{4}}\left( 1+2+3+4+5 \right) $ , Similarly, the result of tens place will also be same just ten times of the ones place. So, add all the places of the digits to get the answer. Complete step-by-step answer: Note: Remember if one of the digits is zero, then all the digits will not have the same cases of appearing in a given place, actually zero cannot appear at the 1000s place, as if it appears at 1000s place, the number will be a 3 digit number. So, in this case you have to fix zero at every place and get the number of times zero is appearing at each place and divide the left out cases among other digits. The given numbers are 1, 2, 3, 4, 5 The total number of arrangements. Using the digits 1, 2, 3, 4 and 5 taking 4 at a time is 5P4 = 5 × 4 × 3 × 2 = 120 ∴ 120 four-digit numbers can be formed using the given 5 digits without repetition. To find the sum of these numbers. We will find the sum of digits at unit’s, ten’s, hundred’s and thousand’s place in all these 120 numbers. Consider the digit in unit’s place. In all these numbers Each of these digits 1, 2, 3, 4, 5 occurs 120 in `120/5` = 24 times in the units place ∴ The sum of the digits at unit’s place = 24(1 + 2 + 3 + 4 + 5) = 24 × 15 = 360 Similarly sum of the digit’s at ten’s place = 360 Sum of the digit’s at hundred’s place = 360 Sum of the digit’s at thousand’s place = 360 ∴ Sum of all four digit numbers formed using the digit’s 1, 2, 3, 4, 5 = 360 × 10° + 360 × 101 + 360 × 102 + 360 × 103 = 360(10° + 101 + 102 + 103) = 360(1 + 10 + 100 + 1000) = 360 × 1111 = 3,99,960
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What is the sum of all 4The sum of all the 4-digit numbers formed using the digits 2, 3, 4, and 5 (without repetition) is 93, 324.
What is the sum of all 4Hence the sum off all 4 diit numbers is 666600. Was this answer helpful?
What is the sum of all four digit numbers whose digits are permutations of 1 2 3 4?Hence, the answer is 259980.
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