What is the sum of all four digit numbers formed by using 1,2 3 4 once in each number by 6?

Since we are going to form four digit numbers, let us have four blanks.

____    ____    ____    ____

The first blank (thousand's place) has four options from the given four digits.

They are 1, 2, 5, 6

If one of the four digits (1, 2, 5, 6) is filled in the first blank, three digits will be remaining.

So, the second place has three options and it can be filled by one of the three digits.

After having filled the second blank, two digits will be remaining.

So, the third place has two options and it can be filled by one of the two digits.

After having filled the third blank, only one digit will be remaining.

So, the fourth blank has only one option and it can be filled by the remaining one digit. 

The above explained stuff can be written as 

                              4 x 3 x 2 x 1  =  24

Therefore the number of four digit numbers formed using  1, 2, 5, 6  is  24.

They are

1256

1265

1526

1562

1625

1652

2156

2165

2516

2561

2615

2651

5126

5162

5216

5261

5612

5621

6125

6152

6215

6251

6512

6521

Is it possible to write all the 24 numbers and find sum of them in exam as done above?

Everyone's answer will be "no".

Then, is there any shortcut ?

Yes. To know the shortcut, come to know the value of "K" using the formula given below.

Here, students may have question.

That is, what do we do with "K" to find sum of all 4 digit numbers formed using 1 2 5 6 ?

Answer is given below.

Concept - Value of "K"

What does "K" do if one of the digits is "zero" ?

Answer :

1. Each one of the non-zero digits will come "K" times at the first place (thousand's place, if it is four digit number).

2. The digit "0" will come "K" times at the second place. The remaining blanks at the second place will be shared equally by the non-zero digits. 

3. The same process which is explained above for the second place will be applied for the third place and fourth place. 

What does "K" do if none of the digits is "zero"?

Answer :

Each one of the non-zero digits will come "K" times at the first place, second place, third place and fourth place. 

How is the above concept applied in our problem ?

In our problem, we have

K  =  24/4

K  =  6

In the given four digits, none of the digits is 0. 

So, each one of the given four digits (1, 2, 5, 6) will come at the thousand's place,hundred's place,ten's place and unit's place 6 ( = K ) times in the 24 numbers formed using 1, 2, 5, 6 (Please see the 18 numbers above).

Sum of Numbers at the First, Second, Third and Fourth Places

To find sum of all 4 digit numbers formed using 1 2 5 6, we have to find the sum of all numbers at first, second, third and fourth places.

Let us find the sum of numbers at the first place (thousand's place).

In the 24 numbers formed, we have each one of the digits (1, 2, 5, 6) six times at the first place, second place, third place and fourth place.

Sum of the numbers at the first place (1000's place)

=  6(1 + 2 + 5 + 6)

=  6 x 14

=  84

Sum of the numbers at the second place (100's place)

=  6(1 + 2 + 5 + 6)

=  6 x 14

=  84

Sum of the numbers at the third place (10's place)

=  6(1 + 2 + 5 + 6)

=  6 x 14

=  84

Sum of the numbers at the fourth place (1's place)

=  6(1 + 2 + 5 + 6)

=  6 x 14

=  84

Sum of All 4 Digit Numbers Formed Using 1, 2, 5, 6 (without repetition)         

Explanation For the Above Calculation :

84 is the sum of numbers at thousand's place. So 84 is multiplied 1000.

84 is the sum of numbers at hundred's place. So 84 is multiplied 100.

84 is the sum of numbers at ten's place. So 84 is multiplied 10.

84 is the sum of numbers at unit's place. So 84 is multiplied 1.

Note : 

The method explained above is not only applicable to find the sum of all 4 digit numbers formed using 1 2 5 6. This same method can be applied to find sum of all 4 digit numbers formed using any four digits in which none of the digits is zero.

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In this answer I presume that all digits $1,2,5,6$ must be present in the number.

There are $4!=24$ such numbers.

We can index them with $i=1,\dots,24$ and write each of them as $$n_{i}=a_{i}+10b_{i}+100c_{i}+1000d_{i}$$ where $\left\{ a_{i},b_{i},c_{i},d_{i}\right\}=\{1,2,5,6\} $

Then $$\sum_{i=1}^{24}n_{i}=\sum_{i=1}^{24}a_{i}+10\sum_{i=1}^{24}b_{i}+100\sum_{i=1}^{24}c_{i}+1000\sum_{i=1}^{24}d_{i}$$

$\left\{ i\mid a_{i}=1\right\} $, $\left\{ i\mid a_{i}=2\right\} $, $\left\{ i\mid a_{i}=5\right\} $ and $\left\{ i\mid a_{i}=6\right\} $ have equal cardinality $\frac{24}4=6$ so: $$\sum_{i=1}^{24}a_{i}=6.1+6.2+6.5+6.6=6(1+2+5+6)=84$$

This can also be applied for $b,c$ and $d$ and finally we find: $$\sum_{i=1}^{24}n_{i}=84+10.84+100.84+1000.84=1111.84=93324$$

Does this really answers your question? If not, then please let me know.