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`        how many natural numbers are their form 1to1000 which have none of their digit repeated.pls give solution`
8 years ago

## Answers : (2)

```							There are 9 one digit numbers
now two digit numbers havng none of their digits repeated are
first digit can not have the digit 0 because then it will be a one digit number which we have already considered. So the total number of possibilities of having the first digit are 9. Now the second digit can have 0 but not that digit which has already be used in the first digit. So total number of possibilities are 9. total possibilities of the two digit numbers having none of their digits repeated are 9*9=81
now three digit numbers are
again the first digit can not be 0 so the total number of possibilities are 9. For the second digit the total number of possibilities are 9(as in the previous case). For the third digit we cannot use the two numbers which we have used for the first two digits. So the total number of possibilities are 8. The total number of possibilities of the three digit numbers having none of their digits repeated are 9*9*8=648
therefore total numbers between 1 and 1000 having none of their digits repeated are 9+81+648=738
```
8 years ago
```							 Total number of possibilities are 9. total possibilities of the two digit numbers having none of their digits repeated are 9×9=81now three digit numbers areagain the first digit can not be 0 so the total number of possibilities are 9. For the second digit the total number of possibilities are 9(as in the previous case). For the third digit we cannot use the two numbers which we have used for the first two digits. So the total number of possibilities are 8. The total number of possibilities of the three digit numbers having none of their digits repeated are 9×9×8=648therefore total numbers between 1 and 1000 having none of their digits repeated are 9+81+648=738
```
5 months ago
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