I always hear about Real Analysis, Algebra, Topology and Number Theory...

Literally disgusting and degenerate. No formal theory behind it.

>tfw never taught combinatorics in high school
>struggle with EVERY SINGLE problem I try to do
>tfw can't even do a simple counting problem

I HATE THIS REEEEEEEEEEEEEEEE

algo.inria.fr/flajolet/Publications/FlSe02.pdf
Learn what a generating function is and how to use it to count.
Learn enough group theory to understand Burnside's Lemma and Polya Enumeration Theorem.

en.wikipedia.org/wiki/Chomsky–Schützenberger_enumeration_theorem

you should be able to solve counting problems with just your IQ you brainlet.

there are humanities students taking graduate exams for fucking management or lawschool who have done virtually no maths since high school who can breeze through counting problems.

there are fucking 14 year olds doing junior IMO papers who've never heard of the pigeon principle or multiplication rule who can work out counting problems.

drop out, you damn brainlet.

Keep trying
youtube.com/watch?v=tYzMYcUty6s
Someone needs some flame retardent

holy fuck i can relate to this

Solve this

There are 7 chairs in a room with 6 people; 3 males, 3 females; a seating arrangement is only considered different when the positions of the people are different relative to each other. A female can never be between two males; what is the total number of possible arrangement? Give answers assuming the chairs are in a row and if the chairs are in a circle.

>solve this
we aren't doing your assignments for you, user

It is the most based field of mathematics, and the only one that will survive the Wildbergerian apocalypse when modern maths eventually crumble upon their illogical foundations

I doubt he can even solve that desu
literally 14 y.o. maths
I wonder... if you can't as well?
mmmm smells like brainlet round here

Not him but I am curious 2480?

It might be easier to take the total number of configurations minus the inadmissible configurations

I got 18 combos for a straight line and defining symmetry as bggbbg = gbbggb

b = boy
g = girl

but I did it with algebra, I dont know much combinatorics formulas

correction, I think its only 6 combos

now im only down to 5.. this is getting weird.

and I realize that this where were at. en.wikipedia.org/wiki/Trace_monoid

The solution is [math] 7 \cdot 4 = 28 [/math] arrangements in a row & [math] 7 [/math] arrangements in a circle.

The logic is that the single empty chair can be at any position.
There is 6 remaining seats occupied by person.
Males always move as a block (with no females in between) so there are 4 possibilities to occupy the 6 remaining seats.
MMMFFF, FMMMFF, FFMMMF, MMMFFF.

In a circle these 4 options become equivalent. Because they become interchange by rotation. (Formal explanation of that by Group Theory)

So my answer is is [math] 7 \cdot 4 = 28 [/math] arrangements in a row & [math] 7 [/math] arrangements in a circle.

I'm a engineering student.

> there are fucking 14 year olds doing junior IMO papers who've never heard of the pigeon principle or multiplication rule who can work out counting problems.

Citation needed.
I formally learned those things in high school, you are usually taught those things if you are in t he math club.

This is completely wrong lol
I can get at least 21 arrangement just by listing on paper
The Males do not need to always move as a block as in:

MFFMMEF

E = Empty
M = Male
F = Female

lol
but anyway Engineers make things happen, you go boyfriend!!!

Hello this thread needs an actual ADULT (combinatorics knower)
this is getting embarrassing

Fucking kek. Niggas wanna diss combinatorics but can't count basic shit.

The question seems a bit ambiguous.
1) Are the people named or un-named?
2) By between do you mean directly between (mfm) or (m_fm) with a gap as well or the even "looser" definition (mfffm) where all three are counted as being between two males?

>named or un-named?
1) I don't think name matters in the question
>A female can never be between two males
Doesn't say explicitly you can't do MEFM or MFEM
And by definition between is 1 between two; among is more than one between two

I gave it the "college try"

Is 3 between 1 and 5? Yes.
3 is also between 2 and 4.
Question seems valid.

between = * 1 *
among = * [math]n[/math] *

OP might need to come back and solve this problem ._.

*u*

>ftfy

>there are fucking 14 year olds doing junior IMO papers who've never heard of the pigeon principle or multiplication rule who can work out counting problems.
>multiplication rule

this fucking guy.... I bet you also say shit like "times it by 10"

Google among
Among def:
1) surrounded by; in the company of.
(disagrees with your usage according to "surrounded by")
2) being a member or members of (a larger set).
(irrelevant to the problem which pertains to ORDERED collections, not UNORDERED SETS)

Kid, I was a decade ahead of where you are when I was your age.

If you have done any serious reading of mathematics utilizing multiple sources, you would know that everybody uses slightly different verbiage when talking about the same objects/concepts. That is why any mathematician worth their salt clearly defines the expressions they use. The devil is in the details. I would rather seek clarification than "ASSume" my internal representation of the construction is the one that is trying to be conveyed.

High school only gets you to the 1600's in terms of parrot-able knowledge but it doesn't teach cleverness or how to prove shit.

Hey bro can you solve it or not

Wouldnt this be easily solved by inclusion/exclusion since its easier to reason about forbidden arrangements? I'm too tired to do it myself/too much of a brainlet.

After drinking some tea, this is my guess for rows:
There are T = 7!/(3!3!) total number of arrangements for the alfabet FFFMMME
There are X = 5!/2! arrangements for the alfabet (MFM)FFME
There are Y = 3! arrangements for the alfabet (MFMFM)FE
So by the principle of inclusion/exclusion it should be equal to T - X + Y = 86 allowed arrangements for the row. Probably wrong if not the circle adds more types of forbidden positions since 86 is not divisble by 7
t. brainlet

Oh yeah, might have forgotten to take into account something when you group them up into (MFM) etc

already did
see

circle is not as simple as divide by 7
Think about mf_fmmf wrapping around.
The f on the right gets surrounded by m's.
I think you would just need to get rid of the shift multipliers for the inadmissibles and divide the total config count by 7.
See

By "positions relative to each other" do you mean that the empty seat is then deleted after meeting the adjacency requirement to yield a string containing 6 symbols?
Might as well add a mirror plane too...