Let's see how many summerfags have invaded this board.
How many 25-bit arrays can you form if they must contain exactly 15 ones and no consecutive zeroes?
8008
>hey /g/ do my homework for me
How about you fuck off.
>>55622705
cant you write somthing that does that?
>>55622705
17 choose 10 is my best guess
Infinini
Permutate and remove duplicates.
Compute the 25! permutations of your arrays and then do a pattern match on them.
Don't post on /g/ until that program finishes running.
>>55622804
DING DING DING
This guy actually knows informatics.
>>55622956
This is 7th grade maths level.
>>55622705
Why 25-bit? Why not take it a step further?
>>55623022
n-bit array then
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1
first zero can go into 16 spots, second zero can go into 15 spots, third can go into 14 spots etc
so this is 16!/6! or
16*15*14*13...*7
i hate counting, but i think this is write
>>55623339
I had the same idea but considering the position of 1's instead of 0's.
For 25-bit the lowest number of 1's that gives no solution is 12.
After that the next 1 can go in 13 places, the next in 12, then 11, then 10.
That gives 13*12*11*10 = 17160. However the 1's are not distinct so this overcounts by n! where n is the number of 1's we used.
We used four 1's so my final answer is 13*12*11*10/4! = 715.
I am off to work so by the time I check this thread again once it will probably be dead. Hopefully someone can find a flaw with this method because I don't think it's correct.
>>55623482
ya im overcounting ofcourse like a retard you are correct
>>55623339
You are off by 10! Zeros aren't unique retard kys
>>55623339
You're permuting the zeroes, you should divide that by 10!
