Question

Find the sum of the four digit numbers formed using the digits$1,2,3,4$
Case 1. When repetition is not allowed.
Case 2. When repetition is allowed.

Answer 1

Hint: For the first case, use the permutation formula$^n{\operatorname{P} _r} = \dfrac{{n!}}{{(n - r)}},0 \leqslant r \leqslant n$to find the count of such numbers. Use the fact that for each place, the number of times each digit appears is the same. Find this number and then multiply it with the sum of all digits and the respective place value. Do it for each place. Then add them to get the answer.
For the second case, there are equal no. of choices for each place. After finding the count, repeat the same steps as in the first case.

Complete step by step solution:
We are given the digits$1,2,3,4$and are asked to find the sum of all the four digit numbers formed out of these digits.
Now, there are two sub-parts to the given question.
The first case is when we will consider only those numbers where not a single digit is repeated.
In the second case we will consider all types of numbers formed using the digits$1,2,3,4$
Let us begin with the first case.
We have 4 digits 1, 2, 3, 4.
There are 4 places where we can place them.
By using the permutation formula, $^n{\operatorname{P} _r} = \dfrac{{n!}}{{(n - r)}},0 \leqslant r \leqslant n$, we know that there are$^4{P_4}$ways of arranging them without repetition.
$^4{P_4} = \dfrac{{4!}}{{(4 - 4)!}} = 4! = 4 \times 3 \times 2 \times 1 = 24$
Thus, there are 24 such numbers.
Consider the unit's place.
Among these 24 numbers, 1, 2, 3, and 4 appear exactly the same number of times.
This means that each digit will appear$\dfrac{{24}}{4} = 6$times
The same situation is repeated for the tens, hundreds, and thousands places.
Let’s find the sum of these 24 numbers.
We need to remember that each digit (1, 2, 3, 4) appears 6 times in each of the places.
Thus, we get the following sum:
\[
   1000 \times 6(1 + 2 + 3 + 4) \\
   + 100 \times 6(1 + 2 + 3 + 4) \\
   + 10 \times 6(1 + 2 + 3 + 4) \\
   \underline { + 1 \times 6(1 + 2 + 3 + 4)} \\
   66660 \\
 \]
Hence, the sum of 4 digit numbers formed out of 1, 2, 3, 4 is 66660 when there is no repetition.

Now, let us consider the case with repetition.
In this case, each of the 4 places has 4 choices.
Therefore, there will be \[4 \times 4 \times 4 \times 4 = 256\]such numbers.
Consider the unit's place.
Now, each digit gets an equal number of chances to appear in the unit's place.
Therefore, each digit will appear $\dfrac{{256}}{4} = 64$times in the units place.
The same situation repeats for the tens, hundreds, and thousands places.
Now, let’s calculate the sum which is done as follows:

\[
   1000 \times 64(1 + 2 + 3 + 4) \\
   + 100 \times 64(1 + 2 + 3 + 4) \\
   + 10 \times 64(1 + 2 + 3 + 4) \\
   \underline { + 1 \times 64(1 + 2 + 3 + 4)} \\
   711040 \\
 \]

Hence the required sum is 711040.
Note: In cases where repetition is not allowed, the formula of permutation is used and not that of combination. In the formula for permutation, the denominator is not divided by r!.