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
How many odd three-digit integers greater than 800 are there
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Lets break the condition into two ranges
801--899
900--999
Case 1:801 - 899
A B C
A = 8 = So definite 1 number
c = 1,3,5,7,9 = 5 possible numbers
B = 10-1-1 = 8
WAYS = 1 X 5C1 X 8C1 = 40
Case 2: 900-999
A B C
A = 9 = 1 POSSIBLE NUMBERS
C = 1,3,5,7 = 4 POSSIBLE NUMBERS
B = 10 - 1 -1 = 8 POSSIBLE NUMBERS
WAYS = 1 X 4C1 X 8C1 = 32
Total Ways = 40+32 = 72
Is the Answer [spoiler]{C}[/spoiler]
801--899
900--999
Case 1:801 - 899
A B C
A = 8 = So definite 1 number
c = 1,3,5,7,9 = 5 possible numbers
B = 10-1-1 = 8
WAYS = 1 X 5C1 X 8C1 = 40
Case 2: 900-999
A B C
A = 9 = 1 POSSIBLE NUMBERS
C = 1,3,5,7 = 4 POSSIBLE NUMBERS
B = 10 - 1 -1 = 8 POSSIBLE NUMBERS
WAYS = 1 X 4C1 X 8C1 = 32
Total Ways = 40+32 = 72
Is the Answer [spoiler]{C}[/spoiler]
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Nice job but this question can be solved by this method.rakeshd347 wrote: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
for hundred place we have 2 options...and for the tens digit we have 9 digit and for unit digit we have 8 places.
So total =9*8*2=144..now out of these 144 half will be even and half will be odd. so answer is C
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Alternate approach:rakeshd347 wrote: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
Good integers = total - bad.
Total:
Number of options for the hundreds place = 2. (8 or 9).
Number of options for the units place = 5. (1, 3, 5, 7, or 9).
Number of options for the tens place = 8. (Any digit 0-9 but the two already selected.)
To combine these options, we multiply:
2*5*8 = 80.
Bad:
Among the integers counted above, a bad integer occurs when both the hundreds digit and the units digit are 9.
P(the hundreds digit is 9) = 1/2. (Since 1 of the 2 options is 9.)
P(the units digit is 9) = 1/5. (Since 1 of the 5 options is 9.).
Thus:
P(both the hundreds digit and the units digit are 9) = 1/2 * 1/5 = 1/10, implying that 1/10 of the counted integers are BAD:
1/10 * 80 = 8.
Good:
Total - bad = 80-8 = 72.
The correct answer is C.
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Hi info2,
Unfortunately, your logic isn't correct - and you got 'lucky' with the approach that you used.
To start, there are actually 5 digits (1, 3, 5, 7 and 9) that could be the last digit. IF the first digit is 9, then there are only 4 options (1, 3, 5 and 7), but if the first digit is 8, then there are 5 options.
Second, there 9 options for the second digit IF you've only considered the first digit. However, there are only 8 options for the second digit when you consider both the first and third digits.
I'd suggest that you take a look at any of the other approaches to see the correct math that could be used to answer this question.
GMAT assassins aren't born, they're made,
Rich
Unfortunately, your logic isn't correct - and you got 'lucky' with the approach that you used.
To start, there are actually 5 digits (1, 3, 5, 7 and 9) that could be the last digit. IF the first digit is 9, then there are only 4 options (1, 3, 5 and 7), but if the first digit is 8, then there are 5 options.
Second, there 9 options for the second digit IF you've only considered the first digit. However, there are only 8 options for the second digit when you consider both the first and third digits.
I'd suggest that you take a look at any of the other approaches to see the correct math that could be used to answer this question.
GMAT assassins aren't born, they're made,
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We need to determine how many integers greater than 800 but less than 1,000 are odd and contain different digits.rakeshd347 wrote: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
Let's start with the numbers from 801 to 899, inclusive.
There is 1 option for the hundreds digit ( the digit of 8), 5 options for the units digit (digits of 1, 3, 5, 7, or 9), and 8 options for the tens digit (since we cannot use the number used in the hundreds or units place. Thus, between 800 and 900, there are 1 x 5 x 8 = 40 possibilities.
Next, let us consider numbers from 901 to 999, inclusive.
There is 1 option for the hundreds digit ( the digit of 9), 4 options for the units digit (digits of 1, 3, 5, or 7), and 8 options for the tens digit (since we cannot use the number used in the hundreds or units place. Thus, between 900 and 1000, there are 1 x 4 x 8 = 32 possibilities.
So, in total, there are 40 + 32 = 72 possibilities.
Answer: C
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Here's a similar one to practice with: https://www.beatthegmat.com/counting-a-3 ... 88550.html
Cheers,
Brent
Cheers,
Brent