find the number of natural numbers from 1 to 1000 having none of their digits repeated?

Hi!

Here is the answer to your question.

 

There are nine one digit numbers from 1 to 9 such that no digit is repeated.

Consider a two digit number whose digit is not repeated.

 

Ones place digit of this number can be filled up using the numbers 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9.

This means the ones place digit is filled in 10 ways

The tens place of this two digit can be filled in 9 ways.

Thus, the number of two digit numbers whose digits are not repeated = 10 × 9 = 90

 

The number of three digit numbers whose digits are not repeated = 10 × 9 × 8 = 720

The number 1000 has repeated digits

Thus, the total number of required numbers from 0 to 1000 having none of digits repeated

= 9 + 90 + 720

= 819

Hope! you got the answer.