SUM OF ALL 4 DIGIT POSITIVE NUMBERS WITH NON ZERO DIGIT
(without repetition)

Number of 4 Digit Positive Numbers with Non-Zero Digit

In all, there are nine different positive non zero digits.

They are

1, 2, 3, 4, 5, 6, 7, 8, 9

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

____ ____ ____ ____

The first blank (thousand's place) has 9 options.(1, 2, 3, 4, 5, 6, 7, 8, 9).

If one of the nine digits is filled in the first blank, eight digits will be remaining.

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

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

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

After having filled the third blank, six digits will be remaining.

So, the fourth blank has six options and it can be filled by one of the six digits.

The above explained stuff can be written as

9 x 8 x 7 x 6 = 3024

Therefore, the number of four digit positive numbers numbers formed using (1, 2, 3, 4, 5, 6, 7, 8, 9) is 3024.

Is it possible to write all the 3024 numbers and find sum of them in exam ?

Definitely, the answer for the above question is 'no'.

Then, is there any shortcut ?

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

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 = 3024/9

K = 336

In total of nine different positive digits, none of the digits is 0.

So, each one of the nine different digits (1, 2, 3, 4, 5, 6, 7, 8, 9) will come at the thousand's place, hundred's place, ten's place and unit's place 336 (= K) times in the 3024 numbers formed using the nine different digits mentioned above.

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

To find sum of all 4 digit positive numbers with non zero digit, 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 3024 numbers formed, we have each one of the digits (1, 2, 3, 4, 5, 6, 7, 8, 9) 336 times at the first place, second place, third place and fourth place.

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

= 336(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 336 x 45

= 15120

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

= 336(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 336 x 45

= 15120

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

= 336(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 336 x 45

= 15120

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

= 336(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 336 x 45

= 15120