Question

Find the sum of all 4-digit numbers that can be formed using digits 0, 2, 5, 7, 8 without repetition?

Answer 1

The given digits are 0, 2, 5, 7, 8

1

2

3

4

The first box can be filled in 4 ways

Using the digits 2, 5, 7, 8 (excluding 0).

The second box can be filled in 4 ways using the digits 0, 2, 5, 7, 8 excluding the digit placed in the first box.

The third box can be filled in 3 ways

Using the digits 0, 2, 5, 7, 8 excluding the digits placed in the first two boxes.

The fourth box can be filled in 2 ways

Using the digits 0, 2, 5, 7, 8 excluding the digits placed in the first three boxes.

∴ Total number of 4-digit numbers = 4 × 4 × 3 × 2 = 96

To find the sum of all these four-digit numbers.

1	2	3	4
			0

Fix the number 0 in the list box (4).

With the remaining numbers 2, 5, 7, 8, box-3 can be filled in 4 ways,

Box-2 can be filled in 3 ways, and box – 1 can be filled in 2 ways.

∴ Total number of 4 digit numbers ending with 0 is = 4 × 3 × 2 = 24 numbers

1	2	3	4
			2

Fix the number 2 in the last box -4.

With the remaining digits 0, 5, 7, 8.

Box-1 can be filled in 3 ways excluding the digit 0.

Box-2 can be filled in 3 ways

Using the digits 0, 5, 7, 8 excluding the digit placed in a box-1.

Box-3 can be filled in 2 ways

Using the digits 0, 5, 7, 8 excluding the digits placed in box-1 and box-2.

Fix the number 2 in the last box -4.

With the remaining digits 0, 5, 7, 8.

Box-1 can be filled in 3 ways excluding the digit 0.

Box-2 can be filled in 3 ways using the digits 0, 5, 7, 8 excluding the digit placed in a box – 1.

Box – 3 can be filled in 2 ways

Using the digits 0, 5, 7, 8 excluding the digits placed in box-1 and box-2.

∴ Total number of 4-digit numbers ending with the digit 2 = 3 × 3 × 2 = 18 numbers

Similarly, Total numbers of 4-digit numbers ending with the digit 5 = 18 numbers

Total number of 4-digit numbers ending with the digit 7 = 18 numbers

Total number of 4-digit numbers ending with the digit 8 = 18 numbers

∴ Total for unit place = (24 × 0) + (18 × 2) + (18× 5) + ( 18 × 7) + ( 18 × 8)

= 18 × (2 + 5 + 7 + 8)

= 18 × 22

= 396

∴ Sum of the digits at the unit place = 396

Similarly Sum of the digits at ten’s place = 396

Sum of the digit’s at hundred’s place = 396

Sum of the digit’s at thousand’s place = 396

∴ Sum of all four digit numbers formed using the digits 0, 2, 5, 7, 8

= 396 × 10° + 396 × 10¹ + 396 × 10² + 396 × 10³

= 396 × (10° + 101 + 102 + 103)

= 396 × (1 + 10 + 100 + 1000)

= 396 × 1111

= 571956