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