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# how many natural numbers are their form 1to1000 which have none of their digit repeated.pls give solution

Fawz Naim
37 Points
11 years ago

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

Yashvanth
15 Points
2 years ago

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