Post by mimisthesc on Feb 13, 2017 10:11:06 GMT 1
I got to this conclussion:
19xx/0x/2x
Where x is a number between 3 and 8 (6 numbers), because the numbers 0,1,2 and 9 are constant, you can't use them anywhere else, because:
The months 01, 02 and 09-12 can't be used, because 1 and 9 are already used in the year, and the 2 is because:
The days between 01 and 22 can't be used because 1 and 9 are already used in the year and the 0 is already used in the months, leaving us the days between 23 and 28, therefore the 2 can't be used anymore in the months.
Back to the formula: 19xx/0x/2x
Since we're talking about combinations without repetitions, we need to use the forumla y = x!
In this case the range of possible numbers is 3-8 (6 numbers) and there are 4 different possible positions (the x's), so we can get to the answer with the following:
Possible days = (6)(5)(4)(3) = 360
In order to be sure what I did is right:
The last two digits [(4)(3)] told me that in each year there were 12 possible combinations, so I can get those combinations manually:
I took the year 1934 for this example. You can use the months 05,06,07 and 08 (because 3 and 4 are used in the year).
When you have 1934/05 the only combinations of days possible are 26, 27 and 28.
When you have 1934/06 the combinations are 25, 27 and 28, and the same happens with the possible months. This tells us that there are 3 possible days per month.
Remember the possible months are 03-08, however, we're using the year 1934, so the possible months would reduce to 05-08 (4 months), where (as said above) only 3 days are possible in each. So in a year, there are 3 possible days in 4 possible months, then, each year has 12 possible days.
If you count every single year that can be done with those rules, you must get the same answer, 360. But that'd take way too much time
Thanks for the problem mimisthesc , it was very fun to solve
Gilberto, good job 
Shall I give you a new one later today?! 
FoIM | 
NNMC | 
OSC
