Solution : We know that the number of 4 digited numbers that can be formed using the digits 1,3,5,7,9 is `""^(5)P_(4)=120`. ltbr We have to find their sum. We first find the sum of the digits in the units place of all the 120 numbers. Put 1 in the units place.
`square square square 1`
The remaining 3 places can be filled with the remaining 4 digits in `""^(4)P_(3)` ways. Which means that there are `""^(4)P_(3)` number of 4 digited numbers with 1 in the units place. Similaly, each of the other digits 3,5,7,9 appears in the units place `""^(4)P_(3)` times. Hence, by adding all these digits of the units place, we get the sum of the digits in the units place.
`""^(4)P_(3)xx1+""^(4)P_(3)xx3+""^(4)P_(3)xx5+""^(4)P_(3)xx7+""^(4)P_(3)xx9`
`=""^(4)P_(3)(1+3+5+7+9)`
`=""^(4)P_(3)(25)`.
Similarly, we get the sum of all digits in 10's place also as `""^(4)P_(3)xx25`. Since it is in 10's place, its value is
`""^(4)P_(3)xx25xx10`.
Like this the values of the sum of the digits in 100's place and 1000's place are respetively,
`""^(4)P_(3)xx25xx1000 and ""^(4)P_(3)xx25xx1000`.
On adding all these sums, we get the sum of all the 4 digited numbers formed by using the digits 1,3,5,7,9. Hence the required sum is
`""^(4)P_(3)xx25xx1+""^(4)P_(3)xx25xx10+""^(4)P_(3)xx25xx100+""^(4)P_(3)xx25xx1000`
`=""^(4)P_(3)xx25xx1111`
`=24xx25xx1111=6,66,600.""^(3)P_(2).` 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:
The first thing we must find out is the total number of possible arrangements of the 4 digits of the given 5, as the number of ways in which 4 out of the given 5 digits can be arranged is the number of 4 digit numbers formed. Therefore, the number of 4 digit numbers formed from the given 5 digits is $ ^{5}{{P}_{4}} $.
So, we are sure that each of the places of the numbers, i.e., ones place, tens place, hundreds place and thousands place have \[\dfrac{^{5}{{P}_{4}}}{5}\] number of times a digit appearing.
So, the sum of ones digit of all the numbers is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right) $
Similarly, tens digit will have the same sum, but we need to multiply the result by 10, as the place value of tens place is ten. So, the value of tens place addition is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10 $
Similarly, the 100s place value is: $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100 $ and 1000s place value is $ \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000. $
Now we will add the place values of all the places to get the answer.
Therefore, the sum of all the numbers satisfying the above condition is:
\[\begin{align}
& \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 10+ \\
& \dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 100+\dfrac{^{5}{{P}_{4}}}{5}\left( 1+2+3+4+5 \right)\times 1000 \\
\end{align}\]
$ =\dfrac{^{5}{{P}_{4}}}{5}\times 15{{+}^{5}}{{P}_{4}}\times 15\times 10+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 100+\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1000 $
$ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\left( 1+10+100+1000 \right) $
$ =\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111 $
Now we know that $ ^{5}{{P}_{4}}=\dfrac{5!}{1!}=120 $ .
\[\dfrac{^{5}{{P}_{4}}}{5}\times 15\times 1111=\dfrac{120}{5}\times 15\times 1111=399960\]
Therefore, the answer to the above question is 399960. 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.
What is the sum of all the 4 digit numbers?
Hence the sum off all 4 diit numbers is 666600. Was this answer helpful?
The sum of all the 4-digit numbers formed using the digits 2, 3, 4, and 5 (without repetition) is 93, 324.
Find the sum of all 4-digit numbers that can be formed using digits 1, 2, 3, 4, and 5 repetitions not allowed? UPLOAD PHOTO AND GET THE ANSWER NOW! Answer : `= 360 + 3600 + 36000 + 360000 = 399960 ` . Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.
So the total number of such 4-digit numbers would be 4x4x4x4 = 256. The average digit in each digit position would equal 2.5 (the average of 1, 2, 3, and 4). So the average number would equal 2.5 * (1000 + 100 + 10 + 1) = 2.5 * 1111. The total would therefore equal 256 * 2.5 * 1111 = 640 * 1111 = 711,040.
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