Bunuel wrote:
Official Solution:
How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 56
C. 72
D. 81
E. 104
In the range 800 - 900:
1 choice for the first digit: 8;
5 choices for the third digit: 1, 3, 5, 7, 9 (since integer must be odd);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
\(1*5*8 = 40\)
In the range 900 - 999:
1 choice for the first digit: 9;
4 choices for the third digit: 1, 3, 5, 7 (9 is out as it's used as the first digit);
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
\(1*4*8 = 32\)
Total: \(40+32 = 72\).
Answer: C
I am confused by the second digit logic.
I went figured 10 figures available to the second digit. And then got rid of any that were included in the two other digits (i.e. for the 800's: I got rid of 8, then all odd digits.. thus 4 remain). This answer got me to 40.
I want to clarify, the reason that my logic was flawed was that I excluded all odd figures in the second digit; thus I excluded far too many numbers. For instance, I would have excluded 871 simply because 7 is an odd number that is capable of appearing in the third digit.
Following this logic, the reason that the second digit has 8 possibilities is because you start with 10 and subtract the 8 (first digit) and odd digit that will appear in the third digit.
Thanks! Want to make sure I am thinking of this correctly.