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Difficulty:
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Question Stats:
63% (02:06) correct
37%
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wrong
based on 1319
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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
M25-20
A. 40
B. 56
C. 72
D. 81
E. 104
M25-20
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23 May 2010, 01:23
Kudos
Bookmarks
How many odd three-digit integers greater than 800 exist, such that all their digits are distinct?
A. 40
B. 60
C. 72
D. 81
E. 104
In the range 800 - 899:
1 choice for the first digit: 8
5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 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 not an option, as it's the first digit)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*4*8 = 32\).
Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).
Answer: C
Hope it's clear.
A. 40
B. 60
C. 72
D. 81
E. 104
In the range 800 - 899:
1 choice for the first digit: 8
5 choices for the third digit: 1, 3, 5, 7, 9 (as it must be an odd number)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 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 not an option, as it's the first digit)
8 choices for the second digit: we have 10 possible digits, but we must exclude the first and third digits, leaving us with 8 digits.
Total for this range: \(1*4*8 = 32\).
Combining both ranges, the total number of odd three-digit integers greater than 800 with distinct digits is: \(40 + 32 = 72\).
Answer: C
Hope it's clear.
28 Nov 2011, 04:25
Kudos
Bookmarks
arjunbt
Yes, you are missing something and there is a reason why Ian did what he did.
Take numbers in the range 500 - 600. How many numbers are there such that all 3 digits are different?
1 (hundred's digit) * 9 (ten's digit) * 8 (unit's digit) = 72
How many of them are odd? 5 is odd so number of possibilities for unit's digit is 4.
[highlight]Number of odd numbers = 1 * 8 * 4 = 32[/highlight]
Number of even numbers = 72 - 32 = 40
Now take numbers in the range 800 - 900.
Total numbers where all digits are different = 72 (as before)
[highlight]Number of odd numbers = 1 * 8 * 5 = 40[/highlight] (now there are 5 possibilities for the unit's digit)
Number of even numbers = 72 - 40 = 32
So the number of even and odd numbers are not half of the total numbers in each range. You see why above. The number of odd numbers depends on the range (whether the hundred's digit is odd or even).
The reason you still got your answer is that the range is 800 - 1000 which has the 800 - 900 and the 900 - 1000 range in it. It averaged out to be half because in the first range 40 numbers are odd and in the second range 32 are odd so they averaged out to 36 i.e. half of 72.
If instead the question says how many 3 digit numbers above 700 are odd with all digits different, the answer will be 104, not 108. Your method will not work in this case. So be careful.
