Question:
Find the number of natural numbers less than (or up to) 1000 having different digits.
Find the number of natural numbers less than (or up to) 1000 having different digits.
Updated On: May 11, 2024
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Solution and Explanation
One Digit numbers = 9
Two Digit numbers = 10P2 - 9P1
= 90 - 9
= 81
Three Digit numbers = 10P3 - 9P2
= 720 - 72
= 648
Therefore , Total = 9 + 81 + 648 = 738
So, the answer is 738.
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