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No problem, things will get better.
Hi,
Yes.
Hi bobbym,
New problem:
There are 2 blue balls, 4 green balls, 8 white balls, 16 red balls and 32 black balls.
What is the probability of picking 5 balls, such that at most two balls of each color is to be included (e.g. [red, red, green, green, black] is fine; but [red, red, red, green, black] is forbidden)
i) Without replacement?
ii) With replacement?
Hi,
Me neither. "concrete mathematics" has one for k=2.
It's bit difficult to find a mathematical solution.
See you later..
Oh, okay.
Hi bobbym,
Partition problems were hard even for Hardy-Ramanujan,
Did you know there is no analytical answer for anything more then a spread of 2?
"spread of 2" means?
Okay, see you later..
Thanks!
Hi bobbym,
Hi bobbym,
Started up raspbian to check the Mm code.
What I thought worked:
Did you nest n-2 times instead of n-1?
For the input I suggested?
Yes, tweak the input: let it be 2, 3, 4 ... 2014 instead of 0, 1 ... 2013
Worked for the python program there.
Their list starts from 0, you took that to consideration?
Hi,
Yes, we need to tweak the code or input.
Anyway, my answer is for your question..
Hi bobbym,
Okay, but I think I have translated your question to algorithm correctly.
Also,
Hi bobbym,
Can I remind everyone that the title of this thread is Easy puzzle challenge. Therefore if your solution is messy and complicated, youre on the wrong tack.
Yet, that's harder than what's shown in #31
Ah, yes!
I should have searched before asking!
Hi,
Yes!
Is there any command in Mm to get moving averages?