Hello i have thought of a easy to solve funny problem. Wanna

>>8604817
Well we'll assume n is x...then x, given a limitational set of correlatives, will be placed next to y, then the number of rotational planes switches back to the same vector

could you define rotational planes please?

>>8604826
OP here. I think you got the right idea, but i dont really understand what you say. You have no Formula so i dont really know.

>>8604835
The number of planes you need to transform from one given coordinate system to another by means of rotation.

bump my dude

how accessible is this to someone who just finished intro to linear algebra which didn't mention rotational planes?

>>8606223
Very accessible and easy.
I give you an example, which can be viewed as spoiler already. The 3 dimensional space has 3 planes, xy yz and xz.

>>8606230
well i accidentally spoiled it for myself when i went on the wiki page for plane of rotation sooo whoops. But I'm going to try to find reason for the formula

>>8606251
the formula on wikipedia is not what i mean there it is n/2.
but for me it actually is described as a sum.
i mean you have a vector in some space and rotate the coordinate system. this rotation can be split into many rotations in every rotational plane. 3 dimensions have 3 rotational planes as i said before. 4 dimensions have 6 rotational planes, 1 dimension has 0 and so on.

>>8606265
they have another definition of plane of rotation i mean it differently.

please don't tell me you're getting at n choose 2 because otherwise kill yourself

>>8606274
n is wrong. my example in 4 dimensions already shows this.

>>8606276
kill yourself.

>>8606281
why ?

>>8606286
let me guess, the 5th dimension has 10 "rotational planes?"

>>8606297
yes

>>8606297
sry if my language confused you im no native speaker.

>>8606298
so basically you mean "The number of distinct 2-d planes formed by the basis vectors of the space"

which is obviously [math]\dbinom{n}{2}[/math] like this guy said >>8606274

>>8606307
yeah i meant that.
i dont understand how the binomial coefficient works, i wrote it as a sum.

>>8606312
share the sum please im interested

>>8606318
sum from i to (n-1) of y_i

>>8606318
now pls explain the binomial ezly.

>>8606324
uh what is y_i defined as

>>8606334
y is a vaiable that is equal to i :D

File: 8de.png (311KB, 680x681px) Image search: [Google]

311KB, 680x681px

>>8606339
fuck

>>8606347
fuck what it gives the correct answer. im too tired to think clear.
now explain binomial plx.

>>8606324
This is obvious. Consider a set of {a, b, c, d, e}

Working from left to right there are 4 combinations of a, then 3 of b from the remaining set, then 2 for c, 1 for d, 0 for e.

The total combinations is just the sum of all the combinations, which in this case would be be the sum of 1 to 4, which is the sum of 1 to n-1.

>>8606358
yeah. but thats not the definition of binomials with explanation but my original answer. OP here.

>>8606361

The binomial (n choose k) gives the number of combinations of k number of elements. n choose 2 gives the number of combinations of 2 elements.

Are you asking why the binomial theorem works for giving the number of combinations?

>>8606365
well thanks now i understand it a bit better. yeah could you explain how that works ?

>>8606373
There are n*(n-1)*...*(n+1-k) ways of choosing elements from a set in some order. But the elements have been counted k! ways, so to get the unique combinations where order doesn't matter you divide by k!


[math]\frac{n\cdot(n-1)\cdots (n-k+1)}{k!} = \frac{n!}{k!(n-k)!}.[/math]

So consider the set {a,b,c}
In this case you have n=3 elements
and you want to know the number of groupings of k=2

So first you list out all the
Permutations (order matters):
{a,b} {b,a} {a,c} {c,a} {b,c} {c,b}

So there are 6 permutations.
But notice that if you don't care about order, then you have counted the number of combinations (where order doesn't matter) 2 times (which is equal to 2!). To get the number of combinations you divide by 2!. 6/2 = 3, which is the number of combinations of two elements in this set.

Try this same principle with a set of 4 elements, but k=3 groupings. You will notice that there are 3! = 6 times as many permutations as combinations.

>>8606399
thanks for dubs and good content.

>>8606399
n = 4 elements
k = 3 groupings

set = {a, b, c, d}

permutations:
{a,b,c} {a,c,b} {b,a,c} {b,c,a} {c,a,b} {c,b,a}
{a,b,d} {a,d,b} {b,a,d} {b,d,a} {d,a,b} {d,b,a}
{a,c,d} {a,d,c} {c,a,d} {c,d,a} {d,a,c} {d,c,a}
{b,c,d} {b,d,c} {c,b,d} {c,d,b} {d,b,c} {d,c,b}

= 24 permutations

But you have counted the number of combinations 3! = 6 times too many times. So you divide 24 by 6 = 4. Therefore there are only 4 combinations of three elements given this set.
Those four combos (in no particular ordering) are:
{a b c}
{a b d}
{a c d}
{b c d}