Given a number n, find how many n digit number can be formed that does not contain 9 as it’s digit.
Input : 1 Output : 8 Explanation : Except 9, all numbers are possible Input : 2 Output : 72 Explanation : Except numbers from 90 - 99 and all two digit numbers that does not end with 9 are possible.
Total numbers of n digit number that can be formed will be 9*10^(n-1) as except first position all digits can be filled with 10 numbers (0-9). If a digit 9 is eliminated from the list then total number of n digit number will be 8*9^(n-1).
Below is the implementation of above idea:
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Improved By : vt_m