Hello i have thought of a easy to solve funny problem. Wanna see how you guys react to it

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

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

kill yourself.

why ?

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

yes

sry if my language confused you im no native speaker.

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

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

share the sum please im interested

sum from i to (n-1) of y_i

now pls explain the binomial ezly.

uh what is y_i defined as

y is a vaiable that is equal to i :D

fuck

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

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.

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

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?

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

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.

thanks for dubs and good content.

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}