1st digit can be any of 2 digits (8 or 9)
2nd digit can be any of 9 digits (anything but the 1 already selected)
3rd digit can be any of 8 digits (anything but the 2 already selected)
So there are 2*9*8 = 144 integers between 800 and 1,000 with three distinct digits.
We're only interested in the odd values, so take out half of them and pick the answer closest to 72. [this is an approximation]
-Patrick
Digits and Probability
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Some test-takers might find it easier just to do a little counting and predict the rest.ruplun wrote:Hi,
how this math problem is solved with proper explanation...
1.How many odd three-digit integers greater than 800 are there such that all their digits are different?
The last digit has to be odd: 1, 3, 5, 7, or 9
The first digit has to be 8 or 9.
The middle digit has to be different from the first and the last.
Say the first digit is 8, the last 1.
For the middle digit we can use anything but 1 or 8, so we have 8 options: 0, 2, 3, 4, 5, 6, 7, 9
So if the first digit is 8 and the last is 1, we can form 8 acceptable numbers: 801, 821, 831, 841, 851, 861, 871, 891.
Since we have 5 options for the last digit, we have 5 * 8 = 40 good numbers in which the first digit is 8.
Using this logic, we'll get another 40 good numbers if the first digit is 9, except that we can't use 9 for the last digit (because then we'd have 2 9's), so we lose 8 of our good numbers. So if the first digit is 9, we'll get only 40-8=32 good numbers.
So there are 40+32 = 72 acceptable numbers.
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ayushiiitm
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Look we have to 3 digits (units, tens hundreds)...
Now greater than 800, we have two options for hundred digit (8 and 9)
if we fill 8 in hundred place then we have 5 options for units place (1,3,5,7,9)
fill one of those place with (1,3,5, 7or9)....you have now 8 options for tens place( 2,4,6,8 or any of 1,3,5,7,9 that was not used in units place)
so total combination = 1*8*5=40
when we take 9 at hundred place
4 options for units place(1,3,5,7)
8 options for tens place (2,4,6,8 or any of 1,3,5,7 that was not used in units place)
so total combination =1*8*4=32
total = 40+32=72
Now greater than 800, we have two options for hundred digit (8 and 9)
if we fill 8 in hundred place then we have 5 options for units place (1,3,5,7,9)
fill one of those place with (1,3,5, 7or9)....you have now 8 options for tens place( 2,4,6,8 or any of 1,3,5,7,9 that was not used in units place)
so total combination = 1*8*5=40
when we take 9 at hundred place
4 options for units place(1,3,5,7)
8 options for tens place (2,4,6,8 or any of 1,3,5,7 that was not used in units place)
so total combination =1*8*4=32
total = 40+32=72
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