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To calculate the 3 digit numbers with exactly one of its digit as 5, we need to use permutation and combination.
Consider a 3 digit number ‘ABC’
Case 1) A=5, thus B and C can be any digits but 5… so total of 9 digits (0,1,2,3,4,6,7,8,9) can replace B and C
Therefore total combinations = 1*9*9 = 81
Case 2) B=5, thus C be any digit but 5 while A can be any digit but 0 & 5. A cannot be 0 as it will violate the rule of being a 3 digit number.
Thus total combinations = 8*1*9 = 72
Case 3) C=5, on similar lines as case 2 thus total combinations = 8*9*1 = 72
Therefore total combinations = 81 + 72 + 72 = 225
Thus, there are 225 3 digit numbers who have exactly one of its digit as 5
Regards
Arun (AskIITians Forum Expert)
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