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Question

Question: How many natural numbers are there from \(1\) to \(1000\) which have none of their digits repeated?...

How many natural numbers are there from 11 to 10001000 which have none of their digits repeated?

Explanation

Solution

We have 3 kinds of digits from 11 and 10001000 such as one digit numbers, two digit numbers and three digit numbers.

Complete step by step answer:

Now here we have to find natural numbers where none of the digits should be repeated.
Here from 11 to 10001000 we have 33 kinds of digits
One digit numbers, two digit numbers and three digit numbers.

One digit numbers:
We know that there are 99 possible to get single digit numbers from 191 - 9
9ways\Rightarrow 9 ways
Two digit numbers:
Here the first digit can be from 191 - 9 and the second digit can be from090 - 9.
We also know that “zero” cannot be the first digit so we have excluded it
Total possible = 9×9=819 \times 9 = 81 ways
Three digit numbers:
Here the first digit can be from 191 - 9 and the second digit can be from 090 - 9 but not the first digit 101=910 - 1 = 9.
And the third digit can be from 090 - 9 but not the same as the first and second digit.
Total possible=9×9×8=6489 \times 9 \times 8 = 648

Here we have found all the possible under without repetition condition
Therefore total number of natural numbers from 11 to 10001000 without repetition= 648+81+9=738648 + 81 + 9 = 738 ways.

Note: Make a note that digits should not be repeated and kindly focus that zero can’t be the first digit for any kind terms.