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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
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