Question

The sum of all five digits numbers that can be formed using the digits 1, 2, 3, 4 and 5 is found. The first digit is (repetition of digits not allowed).

Answer 1

Hint: Here, we will first fix each of the digits at each of the places to find the number of ways the remaining digits can be arranged. Then we will find the sum of the obtained values to find the required value.

Complete step-by-step answer:
We are given that there are five digits five numbers 1, 2, 3, 4 and 5.
If the number 5 is at the unit’s place, the remaining digits 1, 2, 3 and 4 can be arranged without repetition.
We will now find the ways in which the remaining above digits can be arranged.
$4 \times 3 \times 2 \times 1 = 4!$
Thus, there are exactly $4!$ five-digit numbers using the given digits.
So we can now find the ways to be arranged for other digits when any of the given five numbers are at the unit’s place.
We will find the total sum of digits at the unit’s place for all the possible numbers.
$1 \times 4!\left( {5 + 4 + 3 + 2 + 1} \right) = 4!\left( {15} \right){\text{ ......}}\left( 1 \right)$
Similarly if the number 5 is at the ten’s place, the remaining digits 1, 2, 3 and 4 can be arranged without repetition.
Thus, there are exactly $4!$ five-digit numbers using the given digits.
So we can now find the ways to be arranged for other digits when any of the given five numbers are at the ten’s place.
We will find the total sum of digits at the ten’s place for all the possible numbers.
$10 \times 4!\left( {5 + 4 + 3 + 2 + 1} \right) = 4! \times 15 \times 10{\text{ ......}}\left( 2 \right)$
When the number 5 is at the hundred’s place, the remaining digits 1, 2, 3 and 4 can be arranged without repetition.
Thus, there are exactly $4!$ five-digit numbers using the given digits.
So we can now find the ways to be arranged for other digits when any of the given five numbers are at the hundred’s place.
We will find the total sum of digits at the hundred’s place for all the possible numbers.
$100 \times 4!\left( {5 + 4 + 3 + 2 + 1} \right) = 4! \times 15 \times {10^2}{\text{ .....}}\left( 3 \right)$
Also if the number 5 is at the thousand’s place, the remaining digits 1, 2, 3 and 4 can be arranged without repetition.
Thus, there are exactly $4!$ five-digit numbers using the given digits.
So we can now find the ways to be arranged for other digits when any of the given five numbers are at the thousand’s place.
We will find the total sum of digits at the thousand’s place for all the possible numbers.

$1000 \times 4!\left( {5 + 4 + 3 + 2 + 1} \right) = 4! \times 15 \times {10^3}{\text{ ......}}\left( 4 \right)$
Similarly when the number 5 is at the ten thousand’s place, the remaining digits 1, 2, 3 and 4 can be arranged without repetition.
Thus, there are exactly $4!$ five-digit numbers using the given digits.
So we can now find the ways to be arranged for other digits when any of the given five numbers are at the ten thousand’s place.
We will find the total sum of digits at the ten thousand’s place for all the possible numbers.
$10000 \times 4!\left( {5 + 4 + 3 + 2 + 1} \right) = 4! \times 15 \times {10^4}{\text{ ......}}\left( 5 \right)$
We will now find the required sum from the above values for unit’s , hundred’s, thousand’s and ten thousand’s place using the above equation $\left( 1 \right)$, equation $\left( 2 \right)$, equation $\left( 3 \right)$, equation $\left( 4 \right)$, and equation $\left( 5 \right)$.

$$\Rightarrow 4! \times 15 \times \left( {1 + 10 + {{10}^2} + {{10}^3} + {{10}^4}} \right) \\\ \Rightarrow 24 \times 15 \times \left( {1 + 10 + 100 + 1000 + 10000} \right) \\\ \Rightarrow 24 \times 15 \times 11111 \\\ \Rightarrow 3999960 \\\$$

Thus, the first digit in the above value is 3.

Note: In solving this question, we have given 4 digits 1, 2, 3, 4 and 5 it is said that we have to form numbers with the help of these digits. Some students ignore the part of the question where it is mentioned that repetitions of digits are not allowed and solve this question by taking all the 5 digits in all the five cases, that are, $\left( 1 \right)$, $\left( 2 \right)$, $\left( 3 \right)$, $\left( 4 \right)$, and $\left( 5 \right)$.