DAILY MATH CHALLENGE THREAD #2

>>8667766
>How many ways... ...be painted with white, black, red, and blue
infinite? Since a (assuming perfect) sphere has infinite points on its surface which will give you infinite combinations of points containing at least one white, black, red and blue point within the set. Assuming the 'paint' has infinitely small particles.

>>8667834
Each ball is painted one color.

>>8667838
That changes everything.

>>8667766
n := number of balls
colors := {white, black, red, blue}

[math]
\sum_{k=0}^{k=N/2} \binom{n}{2k} \cdot \binom{n}{n - 2k} \cdot \binom{3}{1}
[/math]

[eqn] \sum_{k=0}^{\text{floor}(\frac{n}{2})} {n + 2 - 2 k \choose 2 } [/eqn]

>>8667863
Remember that balls are indistinguishable.

>>8667871
Correct!

>>8667766
Yeah, this sucks, it is not that fun.
We need another problem.

>>8667893
There is still the closed form to be found.

Let E be a set of finite cardinal n.
Find the number of:
-binary relations on E
-reflexive relations on E
-symetric relations on E
-reflexive and symetric relations on E

File: untitled.png (116KB, 1261x541px) Image search: [Google]

116KB, 1261x541px

What is this equation? It does intersect (0.5,1) that's just a graphing error. Yes it is only one equation.

It looks even, discontinuous and periodic. Maybe something defines on the rationals and irrationals.

>>8667954
Two sets of lines. The first intersects x acis at 45/-45 degrees, the second intersects at 30/60.

[math]\frac{1}{48}(45+3(-1)^n+68n+30n^2+4n^3)[/math]

>>8668048
would you care to explain?

>>8668333
nope

each term in the sum in >>8667871 is a degree 2 polynomial in k. You can get the closed form by using well known formulas