find the number of natural numbers from 1 to 1000 having non of their digits repeated.

SOLUTION 

Since we know that natural nos are counting nos and while counting it from 1 to 1000 none of the digits are repeated. Hence the required the required no of natural nos are 1000.

Ans 9(One digit No)+(9*9)(Two digit No)+(9*9*8)(Three digit no)+1(including 1000)

=9+81+648+1

=739

From 1 to 9 we have 9 digitsFrom one to 99 we have 9*9(once digit*tens digit place)=81 digitsFrom 100 to 1000 9*9*8(thousand digit*tens digit*once digit)=648Add 9+81+648=738