
How many natural numbers are there from $1$ to $1000$ which have none of their digits repeated?
Answer
534.3k+ views
Hint: We have 3 kinds of digits from $1$ and $1000$ such as one digit numbers, two digit numbers and three digit numbers.
Complete step by step answer:
Complete step by step answer:
Now here we have to find natural numbers where none of the digits should be repeated.
Here from $1$ to $1000$ we have $3$ kinds of digits
One digit numbers, two digit numbers and three digit numbers.
One digit numbers:
We know that there are $9$ possible to get single digit numbers from $1 - 9$
$ \Rightarrow 9 ways$
Two digit numbers:
Here the first digit can be from $1 - 9$ and the second digit can be from$0 - 9$.
We also know that “zero” cannot be the first digit so we have excluded it
Total possible = $9 \times 9 = 81$ ways
Three digit numbers:
Here the first digit can be from $1 - 9$ and the second digit can be from $0 - 9$ but not the first digit $10 - 1 = 9$.
And the third digit can be from $0 - 9$ but not the same as the first and second digit.
Total possible=$9 \times 9 \times 8 = 648$
Here we have found all the possible under without repetition condition
Therefore total number of natural numbers from $1$ to $1000$ without repetition= $648 + 81 + 9 = 738$ 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.
Here from $1$ to $1000$ we have $3$ kinds of digits
One digit numbers, two digit numbers and three digit numbers.
One digit numbers:
We know that there are $9$ possible to get single digit numbers from $1 - 9$
$ \Rightarrow 9 ways$
Two digit numbers:
Here the first digit can be from $1 - 9$ and the second digit can be from$0 - 9$.
We also know that “zero” cannot be the first digit so we have excluded it
Total possible = $9 \times 9 = 81$ ways
Three digit numbers:
Here the first digit can be from $1 - 9$ and the second digit can be from $0 - 9$ but not the first digit $10 - 1 = 9$.
And the third digit can be from $0 - 9$ but not the same as the first and second digit.
Total possible=$9 \times 9 \times 8 = 648$
Here we have found all the possible under without repetition condition
Therefore total number of natural numbers from $1$ to $1000$ without repetition= $648 + 81 + 9 = 738$ 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.
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