Hi Mayonnai,
The Question asks How many odd three digit numbers greater than 800 are there such that all their digits are different?
i.e 800 -999
According to the explanation given by you & Vinay we have:
8xx: 8 x [1,3,5,7,9] --> the middle [] can be only be [0, 2, 4, 6 ]
if we apply this logic we can say:
first digit is 8, say we take 3rd digit as 1, then according to your logic
i.e 2nd digit should be [0, 2, 4, 6 ]
we can get nos such as
801 , 821, 841, 861
these are correct
what about these nos 831 , 851, 871 ,891
(all these are 3 digit ODD & with all three digits different)
dont they need to be counted as well
You got 40 coz you poeple did the following
8xx: 8 x [1,3,5,7,9] --> the middle [] can be only be [0, 2, 4, 6 ]
now here the last digit can be picked in 5 ways as you stated, but
what does this 5 mean
It can be interpreted as " there are 5 possible digits that can be used as the last digit of the required nos but at a time only one will be used as there is only one position of the last digit"
Your approch assumes that all 5 are used simultaneouly to fill the last position of one nos so we cannot use these 5 as middle hence you shrink the options of the middle nos ie. instead of 8 to just 4. This is incorrect.
The correct approach would be the one used by agps
i.e for 8XX
first digit can be selected in 1 way
last digit in 5 ways [1,3,5,7,9] but not 5 at a time (you have 5 options choose any one say for e.g. 1,)
so we have 8 possibiltes for the middle digit all excluding 8 & 1
so nos ways is 8 X 5 =40
& similarly for 9XX
and correct ams would be 72 only
Regards
Samir
