how many numbers are there between 500 and1000 which have exactly one of their digits as 8?

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Solution

$The required count of numbers can be found using permutation . First fixed the first digit as 8, then One can choose 8 for the first digit in 1 way . One can choose second digit as any of the 9 non eight digits (1, 2, 3, 4, 5, 6, 7, 0, 9) In a similar way, one can choose third digit as any of the 9 non eight digits (1, 2, 3, 4, 5, 6, 7, 0, 9) So the total number of ways are, 1 \times 9 \times 9 = 81 ways Now fixed the second digit as 8, then One can choose 8 for the second digit in 1 way . One can choose first digit as any of the 4 non eight digits (5, 6, 7, 9) In a similar way, one can choose third digit as any of the 9 non eight digits (1, 2, 3, 4, 5, 6, 7, 0, 9) So the total number of ways are, 4 \times 1 \times 9 = 36 ways Now fixed the third digit as 8, then One can choose 8 for the third digit in 1 way . One can choose first digit as any of the 4 non eight digits (5, 6, 7, 9) In a similar way, one can choose second digit as any of the 9 non eight digits (1, 2, 3, 4, 5, 6, 7, 0, 9) So the total number of ways are, 4 \times 9 \times 1 = 36 ways So the total count of numbers between 500 to 1000 which has exactly one of thier digits as 8 are 81 + 36 + 36 = 153$

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