How many natural numbers are there from $1$ to $1000$ which have none of their digits repeated?
Answer
701.7k+ 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.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Onehalf of a convex lens is covered with a black paper class 12 physics CBSE

Differentiate between lanthanoids and actinoids class 12 chemistry CBSE

An object 5 cm in length is held 25 cm away from a class 12 physics CBSE

Trending doubts
Draw a labelled sketch of the human eye class 12 physics CBSE

Which are the Top 10 Largest Countries of the World?

Draw ray diagrams each showing i myopic eye and ii class 12 physics CBSE

Which is the correct genotypic ratio of mendel dihybrid class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

Draw the diagram showing the germination of pollen class 12 biology CBSE

