Question

Find the number of natural numbers from 1 to 1000 having none of their digits repeated.\[\]

Answer 1

Hint: We find the number of one digit numbers without repetition of digits a ${{N}_{1}}$, the number of two digit numbers without repetition of digits ${{N}_{2}}$, the number of three digit numbers without repetition of digits ${{N}_{3}}$, the number of four digit numbers without repetition of digit as${{N}_{4}}$. We find total number of natural numbers having none of their digits repeated as $N={{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}$.\[\]

Complete step by step answer:
We know that from 1 to 1000 there are one digit numbers, two digit numbers, three numbers and four digit numbers. Let us denote the total number of natural numbers from 1 to 1000 having none of their digits repeated be $N$ . Let number of one digit numbers without repetition of digits as${{N}_{1}}$, the number of two digit numbers without repetition of digits as${{N}_{2}}$, the number of three digit numbers without repetition of digits as${{N}_{3}}$, the number of four digit numbers without repetition of digits as${{N}_{4}}$.\[\]
We know that there are total 10 digits in decimal number system 0,1,2,3,4,5,6,7,8,9.There are only 9 numbers with single digit which are the digits themselves 1, 2, 3... 9 excluding the digit 0 in between 1 to 1000.So we have
\[{{N}_{1}}=9\]
We can fill the tenth place of a two digit number with any digit excluding 0 which we can do in $10-1=9$ numbers of ways. We can fill the unit place with any of the 10 digits excluding the digit we have already filled in the tenth place to avoid repetition which we can do in $10-1=9$ number of ways. So by the rule of product
\[{{N}_{2}}=9\times 9=81\]
We can fill the hundredth place of a three digit number with any digit excluding 0 which we can do in $10-1=9$ numbers of ways. We can fill the tenth place of the three digit number with any digit excluding the digit we have already filled in the tenth place to avoid repetition which we can do in $10-1=9$ number of ways. We can fill the unit place of the three digit number with any digit excluding the digit we have already filled in the tenth and hundredth place to avoid repetition which we can do in $10-1-1=8$ number of ways. So by the rue of product we have,
\[{{N}_{3}}=9\times 9\times 8=648\]
There is only one four digit number in between 1 and 1000 that is 1000 itself which has the digit 0 repeated 3 times. So we have
\[{{N}_{4}}=0\]
So by the rule of sum we have,
\[N={{N}_{1}}+{{N}_{2}}+{{N}_{3}}+{{N}_{4}}=9+81+648+0=738\]

Note: We can alternatively solve by negation method where we find the number of numbers with digit repeated and then subtract from total digits 1000. If there are $m$ number of ways to do one thing $n$ number of ways to do another thing we can do either of them in $m+n$ by the rule of sum, and both the things in $m\times n$ ways by the rule of product.