Question

The sum of all five digit numbers that can be formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is
(1) \[366000\]
(2) \[660000\]
(3) \[360000\]
(4) \[3999960\]

Answer 1

Hint: We are given that five digit numbers using the numbers 1, 2, 3, 4 and 5 are formed with no repetition and we are asked to find the sum of those numbers. We will find the sum of each of the unit’s place digits, ten’s place, hundreds place, thousand’s place and ten thousand’s place digits and then add them up to find the total sum of the numbers.

Complete step by step answer:
According to the given question, we are given that five digit numbers using the numbers 1, 2, 3, 4 and 5 are formed with no repetition and we have to find the sum of those numbers.
We can find the sum of the digits of the unit’s place digits, ten’s place, hundreds place, thousand’s place and ten thousand’s place and then we can add them up to find the total sum of the numbers formed.
For unit’s place –
If 5 is at the unit’s place, then the remaining digits can be arranged in \[4!\] ways.
Total sum of digits at unit’s place for all number is \[4!\left( 1+2+3+4+5 \right)\] ----(1)
For ten’s place –
Total sum of digits at ten’s place for all number is \[10\times 4!\left( 1+2+3+4+5 \right)\] ----(2)
For hundreds place –
Total sum of digits at hundreds places for all numbers is \[{{10}^{2}}\times 4!\left( 1+2+3+4+5 \right)\] ----(3)
For thousand’s place –
Total sum of digits at thousand’s place for all number is \[{{10}^{3}}\times 4!\left( 1+2+3+4+5 \right)\] ----(4)
For ten thousand’s place –
Total sum of digits at ten thousand’s place for all number is \[{{10}^{4}}\times 4!\left( 1+2+3+4+5 \right)\] ----(5)
In order to find the total sum of all the numbers possible, we will add up the sums of the digits we calculated in previous steps, we get,
 Adding up the equations (1), (2), (3), (4) and (5), we get the expression as,
\[4!\left( 1+2+3+4+5 \right)\left( 1+10+{{10}^{2}}+{{10}^{3}}+{{10}^{4}} \right)\]
\[\begin{align}
  & \Rightarrow 24\times 15\times \left( 1+10+{{10}^{2}}+{{10}^{3}}+{{10}^{4}} \right) \\
 & \Rightarrow 24\times 15\times \left( 11111 \right) \\
 & \Rightarrow 3999960 \\
\end{align}\]

So, the correct answer is “Option 4”.

Note: While adding up the terms make sure that no terms are left else that might lead to wrong answers. The sum of the possible five digit numbers can be also found by using the formula:
\[\left( Sum\text{ }of\text{ }all\text{ }digits \right)\left( n-1 \right)!\left( \dfrac{{{10}^{n}}-1}{10-1} \right)\]