Re: How many even 3 digit integers greater than 700 with distinct non zero
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24 Nov 2021, 10:21
Notice that we are told that no digit in these numbers could be zero. So, we have only 9 digits to use for XYZ, where Z is an even number.
If the first digit (X) is 7 or 9 (2 values) then the third digit (Z) can take all 4 values (2, 4, 6, or 8) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 2*4*7=56 numbers;
If the first digit (X) is 8 (1 value) then the third digit (Z) can take only 3 values (2, 4, or 6,) and the second digit (Y) can take 7 values (9 minus two digits we already used). So, for this case we have 1*3*7=21 numbers;
Total: 56+21=77 numbers.
Answer: E.