Bunuel
Official Solution:
How many odd three-digit integers greater than 800 are there such that all their digits are different?
A. 40
B. 60
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
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
Total for this range: \(1*5*8 = 40\).
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 the first digit)
8 choices for the second digit: 10 digits - first digit - third digit = 8 digits.
Total for this range: \(1*4*8 = 32\).
Total: \(40+32 = 72\).
Answer: C
Hi Bunuel,
I just wanted to clarify this question, as i didn't understand it properly. If the question is asking how many three digit
odd integers then why are we including the numbers from 801 - 899...shouldn't these numbers not be counted because these numbers have an 8 in them? (even number) . I thought the question is asking the possibilities of 3 digit odd integers that are all different..example 913 - all odd and the digits are different. I hope my question makes sense, i'm just a bit confused. Thank you.