1:19 AM
Are there any Boolean algebra experts around?

1:40 AM
When trying to show that a group is free abelian, it's sufficient to show that it has a basis, right? Why is that enough?

1:56 AM
@Rithaniel: I haven't thought about this in a long time, but it depends on your definition of basis, I guess. You need to be sure there are no relations among your generators.

Well, the definition we've been using is that a basis for a free abelian group is a set that generates the group, and that for any finite subset $\{x_i\}_{i\in I}$ of this generating set, $\sum\limits_{i\in I}n_ix_i=0$ with $n_i\in\mathbb{Z}$ implies that each $n_i=0$.
I should look up some proofs this weekend.

2:34 AM
Right. That's saying no relations.

1 hour later…
3:45 AM
@Rithaniel Heh, that's almost like "linearly independent" with vector spaces
except that, in $\Bbb Z_n$, every singleton (set with one element) is linearly dependent
because $nx=0$ but $x\ne0$
(Well, I suppose in a vector space, $\{0\}$ is linearly dependent.)
(So it's not so strange.)
(Just strange that non-identity elements could do that.)

Yeah, basically, saying that a group has a nonempty basis is equivalent to saying that it's isomorphic to the direct sum of $x$ many copies of $\mathbb{Z}$
So, I suppose it might be the correct spirit to say that it's the same thing as a vector space, but instead of being over a field, it's over $\mathbb{Z}$.

Yeah
The word is "module"
A module is basically a vector space over a ring instead of a field

Oh yeah, I've heard of those. It's a vector space, but for-- ninja'd.

Yeah so an abelian group is the same concept as a module over $\Bbb Z$
I don't know how bases work for modules in general though
All I know is the word

4:00 AM
Well, the definition of a basis we've been using seems to force the object the basis if for to specifically be a free abelian group.

I've heard the phrase "Grobner basis"
It might be related, I dunno
I don't actually know what that is
but it has 'basis' in it so it might be related

Right now, my brain is on the topic of infinitely generated abelian groups that are not free abelian groups. I want to say that $\mathbb{Q}$ under addition is an example.

Makes sense
Everything can be divided by two, so whatever you have in your basis can be written as the sum of smaller things

Apparently, if you have that $G$ is abelian, finitely generated and, each element has infinite order, then $G$ is free abelian. However, if $G$ is abelian, not finitely generated, and each element has infinite order, then it isn't free abelian. I'm trying to exactly hone in on what is changing between the two.
(I mean, obviously what is changing is that you need infinitely more generators for one than the other, but I mean in a more philosophical sense)

If @Secret were here he'd ask what if the generating set were Dedekind-finite

4:09 AM
Anyways, what've you been up to, lately, Akiva?

A bunch of Japanese assignments for next week were posted online
so I'm trying to get them all in at once
using the textbook

That's how you do it. This HW I'm working on now isn't due until Friday.

There's a test on Monday

Ah, in Japanese?

@Rithaniel Yeah
Gotta know those charts^ plus some vocab and grammar
Every syllable has two characters

4:12 AM
I wish you luck on it. I'm terrible when trying to learn other languages.

and there's not much similarity between the two syllabaries

Well, Shi and Tsu are currently my favorites.

Maybe better picture ('cause they're side-by-side)
@Rithaniel シ and ツ?

Yes, the faces

ソ and ン (so and n) are great too
because they're essentially identical
and make you struggle for half a minute trying to read パソコン
(It's pasokon - short for 'perso'nal 'com'puter)
(ハ ha plus circle equals パ pa)

4:17 AM
I wonder if people learning English have the same sort of issues with P, B, and R.

b d p and q are the same letter
If <q> were pronounced /t/ it would be so much better
because then we'd be able to say "stem up = voiced, stem down = voiceless"
But no <q> ruins the pattern

Just rotated $\pi$ radians or reflected about the horizontal axis.

Rotating 180 degrees is a point reflection about the origin
b <-> q
p <-> d
not a reflection across an axis

Ah right, should be horizontal reflection
b <-> p
q <-> d

Fun fact: the Japanese word for bread is 'pan' - a Portuguese loanword
because there is no wheat natively in East Asia
(pretty sure anyway)
The staple crop is rice
(The word for "meal" also is the same as the word for "white rice", incidentally)
(Kinda like lechem in Biblical Hebrew meaning both "bread" and "food")
Nice when little cultural things come through in the language like that

4:23 AM
Yeah, it's interesting how culture affects language

And both are malleable
Culture changes over time
and with it, language

I wonder how much the more strange traits of languages come from culture. Like phonetic inconsistency, as an example.

Phonetic inconsistency?

Like through and bough

Oh
I would call that orthographic inconsistency
I guess it depends on your perspective
Recently saw this thing: zompist.com/yingzi/yingzi.htm
"If English was written like Chinese"

4:28 AM
Well that's an interesting thought experiment

> the volve in revolve, evolve, involve, devolve, which would have its own yingzi, and would seem as much a "word" or component of the language as the match in rematch, mismatch, unmatch
> Worse yet, the -cuit of biscuit and circuit might be written with the same character (a derivative of kit), and a meaning sought for it-- perhaps 'round', since biscuits are round and circuits involve going round. Again, etymologically this is nonsense.
> There would even be a tendency to think of words as derived from characters rather than the other way around.

@AkivaWeinberger much more interestingly is the fact that tempura is also a loanword

Is it really?

Perhaps the words ending in "-morphism" would end in "-volve"

> From Portuguese, ultimately from Latin. Different dictionaries link two different original terms:

> Portuguese tempero (“seasoning”) or tempera (“he/she/it seasons; season!”), third-person present singular or imperative tense of temperar (“to season, to temper”), from Latin temperare (“to mix, to temper”).[1][2][3]
> Portuguese têmpora (“Ember days”), from Latin tempora, plural of tempus (“time; period”). When Portuguese explorers (mostly Jesuit missionaries) arrived in Japan, they abstained from eating beef, pork and poultry during the Ember days series of holidays. Instead, they ate fr

4:30 AM
temperar means to season but also sometimes to cook

Interesting

Hi, @Eric, DogAteMy, re @Rithaniel.

Heya Ted

@Rithaniel: Relative to your earlier comment, you might want to investigate the difference between direct sum and direct product ...

4:45 AM
Ooooo, yeah. So if you have an abelian, infinitely generated group such that each element has infinite order, it might be the direct product of infinitely many copies of $\mathbb{Z}$?

5:10 AM
I get way too interested sometimes in where names of math things come from
case in point, my comments here: math.stackexchange.com/questions/3363419/…

2 hours later…
7:16 AM
Tautology club meets at least twice a week, if not less

@AkivaWeinberger well if it isn't less than twice a week then it must be at least twice a week!

7:49 AM
Can I show without using Zorn's lemma that a ring is local iff it's non units form an ideal?

8:22 AM
The dream of every cranks ever
"That someone with almost no mathematical training end up stumbled upon famous problems and solutions"

8:35 AM
It's sometimes interesting when one investigate gaps in literature. For something as simple as parabolas, there is a lot yet to be understood at the geometric level. The paper in this work finds a differential equation that governs the existence of the foci, and form that derived the parabola as the only solution
While geometric proofers and and proof verifiers do exists, they are still less advanced compared to algebraic proofers. It does makes people wonder, just how much secrets were still hidden in the set $\Bbb{F}^2$ to be discovered

1 hour later…
9:55 AM
Morning

10:38 AM
Morning
What do you people think of my interpretation of groups? math.stackexchange.com/questions/3365408/…

1 hour later…
11:49 AM
If X*X is hausdoff then is X hausdorff ?

12:39 PM
@ABStkokes No in general. Consider the space $A = \{a,b,c\}$ with the topology $\Phi = (\{a,b\},\{c\},A,\phi)$. This is not hausdorff as $a,b$ can't be separated but look at the product $A \times A$ with the topology being given by the usual product topology.

12:59 PM
@AkivaWeinberger Hmm... I am guessing assuming axiom of choice, that the group is not finitely generated will mean there exists at least one element $g$ from $\Bbb{Z}^{\aleph_{\alpha}}$ for some $\alpha \in \text{On}$ which no finite subset of $G$ can express $g$ as a linear sum, hence there is no (finite) basis for $G$ and hence it cannot be free abelian

Nvm, that's wrong

Assuming choiceless universe, suppose that there is one element $h$ from $\Bbb{Z}^{\Delta}$ where $|D|=\Delta$ is Dedekind finite. Then it still seemed to be true that there are no finite sets in $G$ which its span can produce $h$ as the finite Cartesian product of finite sets is still finite, so any finite linear combination of elements in $\Bbb{Z}^n$ for $n \in \Bbb{N}$ will have to miss out $h$, and hence $G$ will fail to have a basis

Hey @Secret
o.o I made another ordinal collapsing function based on Greatly Mahlos

I am not sure, but it seems there is a well ordered countable subset isomorphic to $\Bbb{N}$ in $\Bbb{Z}^{\Delta}$ (just pick the constant sequences $(0,...),(1,...),(2,...),...$ and inject this to $\Bbb{N}$), thus I think one does not even need the axiom of choice to prove that statement of Rithanial
@SimplyBeautifulArt will check shortly

### This is the Realm of Simply Beautiful…

Room for totally bored people to hang. Open discussions.
simpleart: post it when you are ready

1:18 PM
@ABStkokes argue like this, if X*X is hausdorff then the diagonal is closed. Hence X is hausdorff.

2:00 PM
@ABStkokes Consider two points with the same x-coordinate
@Secret The whole point is we want to know if it looks like $\Bbb Z^X$ or not
If it's finitely generated then yes, if not then $\Bbb Q$ is a counterexample
(Given abelian and everything has infinite order)

What's the question exactly? @Akiva

Heya @Alessandro

Hi @ÍgjøgnumMeg

How's Italien?

   @SayanChattopadhyay     There 4 circle how many maximum  possible way they can intersect answer is 12,formula 4c2

2:08 PM

I move in 1 week!
Taking any nice courses this semester?

@SayanChattopadhyay but I am not able think how this formula nc2 work

I'm going to do Models of Set Theory II, a set theory seminar, global anaysis I and another course but I'm not sure yet which one
I heard you're going to do a lot of number theory

Nice, I'm sure that'll be fun for you (I'd die)
And yeah I'm doing ANT I, modular forms, L-functions, and a topology proseminar (because i'm a noob in topology)

I'm starting to prepare my seminar talk for set theory and I understand very little of what's going on

2:10 PM
and a short minicourse on iwasawa theory for elliptic curves for the first two weeks lol

I need to find someone who knows this stuff to ask a couple questions

@AlessandroCodenotti If you have a finitely generated abelian group such that every element has infinite order, it looks like $\Bbb Z^X$.

@ÍgjøgnumMeg What's a proseminar?

If it's infinitely generated, it might not, for example $\Bbb Q$.
What if it's Dedekind-finitely generated?

2:11 PM
It's like.. a mini seminar for basic stuff I guess, I don't think it counts towards my masters

Ah I see

just doing it so that I have some background in topology lol, my talk will be on covering spaces

@yuvrajsingh A pair of circles can only intersect at two points, right?
There are six pairs (AB, AC, AD, BC, BD, CD), and each generates a maximum of two intersection points

@ÍgjøgnumMeg That's a useful topic to know (even in number theory)

Yeah :) I will see if I can work in some of Grothendieck's Galois theory to the talk
@Alessandro what will your talk be on?

2:19 PM
Do you know Fubini's theorem?

I know of it, I guess

So you take a measurable function $f:\Bbb R\times\Bbb R\to [0,\infty)$ and you get that the two iterated integrals are equal and they are also equal to the integral on the product

yis

2:23 PM
(you can do this with $\sigma$-finite measure spaces, but I only care about real functions for the talk)
The question is whether the hypothesis can be relaxed, instead of asking that $f$ is measurable let's just take $f\colon\Bbb R\times\Bbb R\to[0,\infty)$ such that for almost all $x$ the functions $f_x$ and $f^x$ are measurable (those are obtained from $f$ by fixing respectively the first and second argument to $x$) and such that $x\mapsto \lambda(\int f_x)$ and $x\mapsto\lambda(\int f^x)$ are both measurable (those all follow from measurability of $f$ in the usual version).
Must the two iterated integrals be equal?
And this strong version of the theorem turns out to be independent of $\mathsf{ZFC}$, in particular my talk will be on the construction of a model in which it holds

@AkivaWeinberger this is when you take two at a time

@Alessandro sounds cool!
Totally above me but I can understand why it would be interesting lol

Every talk of the seminar is going to be about a problem not necessarily in set theory which turns out to be independent

that's a nice idea

There's going to be a couple of talks on whitehead's problem as well for example (suppose that $A$ is an abelian group with $\mathrm{Ext}^1(A,\Bbb Z)=0$, is $A$ free?)

2:31 PM
I mean when I take 3 circle, because the question ask maximum, intersection

@Alessandro which (I googled it) is undecidable in ZFC?

cool
undecidability seems like such a minefield
@Alessandro I found a nice block of flats in Mannheim to live in and signed my contract this week
which means I have 2x20km cycles every day lol

That's still better than sleeping under a bridge :P

Correct
hahah

2:56 PM
Trying to compute the set of all infinitely generated abelian group where every element have infinite order and is not free is basically going nowhere
this is because the linear dependent relation can be practically any finite linear combination
$\sum_{x \in X}nx = y$
Actually, does a divisible abelian group implies it is not free abelian?
> In contrast, non-zero free abelian groups are never divisible, because it is impossible for any of their basis elements to be nontrivial integer multiples of other elements
and that seemed to have nothing to do with cardinality

3 hours later…
5:56 PM
If a group acts freely on a connected graph, does it mean it acts properly discontinuously?

6:19 PM

6:32 PM
@yuvrajsingh An extreme case would be if they are concentric and of the same radius, then they intersect at infinitely many points. Leaving this stupid case aside, do as Akiva says.

@user193319 I think so

That would be amazing. Because then I can apply the Svarc-Milnor lemma to this problem I'm working on.

1 hour later…
7:43 PM
Hmm I'm not sure thinking again about it

1 hour later…
8:53 PM
How can I prove that the tangent to a point on a hyperbola intersects the perpendicular line passing through the focus on the auxiliary circle?

9:50 PM
Having some difficulty. I have that $\sum\limits_{i=2}^kn_ix_i=-n_1x_1$ with $x_i$ in some group $G$ and $n_i\in\mathbb{Z}$. How would I go from here to the assertion that either all $n_i=0$ or $G$ has an element with finite order?

@Rithaniel augmentation map?

Is that supposed to be an additive group? Otherwise I can't figure out what $x_1+x_2$ is supposed to mean

This is in $\Bbb Z[G]$ I guess
augmentation sends $\sum_{i=1}^k n_ix_i \mapsto \sum_{i=1}^k x_i$

Yeah, abelian group so we've been using additive notation.

fair enough

9:54 PM
and $0 \in \ker \varepsilon$

So, what is the context for the augmentation map?

it's a map $\varepsilon : \Bbb Z[G] \to G$ that sends $\sum_{g\in G} n_ga_g \mapsto \sum_{g \in G}a_g$ where $n_g \in \Bbb Z$ and $a_g \in G$
if you're summing over a finite number of elements then the relation above gives you an element of finite order in $G$ (namely $\sum_{i=1}^k x_i$) because $0 \mapsto 0 = \sum_{i=1}^k x_i$
I guess

Okay, so I assume this map is not 1-1?

uhh I don't think so
because it's a homomorphism I think
this might be bullshit

Yeah, I'm having trouble figuring it out, to be honest.

10:01 PM
yeah the augmentation map goes to Z not to G
ignore
hahahaa

Ah, darn, it seemed promising.

alas, I was misremembering smth I read ages ago

So, if I could somehow even out the integer coefficients, this would be easy to show.
I don't think that's going to be possible, though.
So $(n_1-1)x_1+\sum\limits_{i=2}^kn_ix_i=x_1$ and so $(n_1-1)x_1=(n_1-1)x_1+\sum\limits_{i=2}^kn_ix_i-\sum\limits_{i=2}^kn_ix_i=x_1-n_1x_1=(1-n_1)x_1$

@Rithaniel: What if you take the lcm of all your $n_i$?

So, $(2n_1-2)x_1=0$? Have I made any mistakes?

10:13 PM
Oh, I'm going back to the original relation. But I'm still not right.

Hey @Ted :)

Hi @ÍgjøgnumMeg.

Wait, I did make a mistake, I think.
Also, hey Ted

hi! help me understand the null cone in conformal geometry? (in particular I'm most familiar with the "conformal geometric algebra" construction)

I thought about multiplying everything by the lcm of the $n_i$, but part of my brain was insisting that I would run into issues with that.

10:16 PM
I guess we can think about the way we'd get to the Smith normal form using the relations. Suppose you have $\Bbb Z^3$ and the generators satisfy $2x+y-z=0$. Do you know what group that is isomorphic to? Oh, so it's going to depend on the relation. If the relation turns into $x=0$, then you still have a free abelian group, just of lower rank. So I don't believe your assertion.
You need to know more about your $n_i$. If they have a gcd $>1$, that'll do.
But if they're all relatively prime, you can just get rid of one of the generators.
@Rithaniel: You thinking?

Yeah, bouncing ideas around in my head.

So, for example, with the one relation I gave you, you don't get any torsion. You just get rid of one generator.

(I should probably mention that I don't know what Smith normal form is.)

You basically put — in this case — a matrix with integer entries into diagonal form, using row and column operations (over $\Bbb Z$). It's super powerful and useful.
You can read about it in Artin's Algebra (although he doesn't name it) or zillions of other places.

7 days to go!

10:27 PM
But who's counting ...

... cough

I told you to quit that nasty smoking~!

hahaha I will actually quit when I leave the UK
I'm excited for my 40km commute

@Rithaniel: In your algebraic "manipulation" it's nonsense. Your first equation is off by a sign.

I'm looking at this: https://en.wikipedia.org/wiki/Conformal_geometric_algebra#Mapping_between_the_base_space_and_the_representation_space

Am I understanding correctly that any (hyper)-parabolic section of the null cone contains representations of every point in the base space?

10:31 PM
Yeah, wanted to delete it because I realize I made a mistake, but I was too slow. Perhaps my head's not in the right space for math, tonight.

Well, I still say the result is false without further hypotheses. Is this printed in a book somewhere?

It's a little confusing to me that the parabola is called "horospheres"

When you projectivize (include points at infinity), a parabola looks just like an ellipse or circle. Presumably that thing only looks 1-dimensional in this picture, but it's really higher dimensional. I don't want to think about it.

And I've never learned geometric algebra.

10:35 PM
That's helpful @TedShifrin! That's one of those things that I've seen insinuated many times, but never heard anyone say it outright.

Same for a hyperbola, by the way. What I'm saying is special about curves in $\Bbb P^2$. In $\Bbb P^3$, nonsingular quadric surfaces can be of two different topological types (working over $\Bbb R$).

Okay... so this cartoon makes it seem as though any conic section of the null cone could be used as a complete representation of the points in the base space.

That's what "projective" geometry does for you. All slices of the cone are projectively equivalent.

hmm, very intriguing!

You can actually see that discussed in the last chapter of my linear algebra book (along with applications of linear algebra to computer graphics).

10:41 PM
So in the conformal geometric algebra, there are operations that take base space point representations (null vectors on that cone) to other base space point representations. I'm understanding now that basically all those "point to point" transformations are just maps from the null cone to itself. Pretty cool!
p.s. I'm searching for your book now :D

There's a link on my homepage, linked in my profile. Not that I'm twisting arms :P

I'm having trouble with a combinations problem. The problem is: There are 4 people who need an outfit built for them at a department store. The outfits are built in 3 steps. First the pants is chosen, then a shirt, then shoes. The department store can start at any person but must build them always in that order: pants, shirt, then shoes. If each person already knows what he/she wants, how many ways can these outfits be built.
For example, the store can first assign person 1 their pants, then assign person 1 their shirt and then finally his shoes before moving on to person 2 and repeating. Or the store can assign person 1 pants, then assign person 2 pants, then assign person 2 shirt, then assign person 1 shirt, then person 3 pants, etc.

@TedShifrin I see you also have a differential algebra book. Out of curiosity, any reason you haven't studied geometric algebra?

Nothing with differential algebra.
Plain old abstract algebra ... and differential geometry.

I tried going about this and thinking that for the first step: the store has 4 options (pants for any of the 4 people). Then after that they have 4 options again (shirt for whoever was first chosen or pants for any of the other 3 people). But using this approach the next step is always dependent on the last

10:45 PM
oops, that's what I meant

So I get stuck

Geometric algebra seems very popular among a small group of people (mostly physics-oriented), and I stuck with differential forms. They were all I needed.

That's an undergraduate differential geometry text, @micahscopes. Just for regular old people who know multivariable calculus and basic linear algebra.
Stuck trying to do what, @krauser?
Oh, I see you put the problem up above.

Yea, have a look at it :) Thank you

10:47 PM
The pants, shirt, and shoes are all independent of one another. No one says the outfits have to look good.
But I don't understand what "If each person already knows what he/she wants" means.
If I've decided I want a specific outfit, then I don't have any choices.
The problem is way too vague. Does the store have exactly 4 pairs of pants, all different? Same for the shirts and shoes.
And what exactly are we trying to count? All possible outfits for the 4 people?

What I mean is that the 4 people have already decided their outfits, but the department store still needs to give each of the 4 people what they wanted. The question is asking, how many ways can the store give these people their outfits while still adhering to the rule that they must give a person pants first, then shirt, then shoes

Sorry. The problem makes no sense to me for the reasons I already specified.

The store as exactly 4 of each item, all distinct

The order makes no difference. I don't understand why they're emphasizing that.

Yea it does

10:51 PM
OK, so if you and I both want brown pants, the store can't make us both happy.
Why does the order matter?

No no. The store has the exact quantity of what we each want.
Its just a matter of how many different orders can the store give out the materials to the people
So if theres just two people, the store could either do: [Pants1 , Shirt1, Shoes1, Pants2, Shirt2, Shoes2] or [Pants1, Pants2, Shirt1, Shirt2, Shoes2, Shoes1] or [Pants2, Shirt2, Pants1, Shoes2, Shirt1, Shoes1] , etc.

So we don't care about what the colors are. The question is asking about different ways of putting things in order.

These are all different orders that he store can give out what we each want, while sticking to the rule that they can only give a person pants before a shirt, and a shirt before shoes
Yea
The style of each piece is already fixed for each person
Each person already knows what he/she wants and is always going to get that item

@TedShifrin thanks for the chat! I hope you find a chance to dip your toes into Clifford algebra some time it's great.
(or geometric algebra, which is nice for applications)

I know something about Clifford algebras, @micahscopes, but from a different perspective. But I no longer have my library with resources on them ...
As I said, I've stuck with differential forms for my career and that's good enough :)

10:56 PM
you do you ;)

OK, @krauser, so I get the point finally.
So the rule is that P1, P2, P3, P4 can go anywhere in our long list, but S1 must follow P1 and T1 must follow S1. Same for the others.
(I'm using S for shirt and T for shoes.)
This seems like a complicated version of bars and stripes.
@krauser: So let's try to do it with just two customers.
I think you want to think of 6 slots and assign 3 of them to the first customer and the remaining 3 to the second customer. Obviously, the first customer determines everything in this case. How many ways can we do it?
@krauser: Are you here?

Sorry yea was trying to solve it

You follow my last suggestion? I think that's the way to do it.

So to think of it as just all the materials in a basket and we are assigning 3 to each person?
I was thinking that. Like having (12C3)(9C3)(6C3)(3C3)

Well, I really want you to think of the slots, because that determines the order.
There you go. That's the right answer. Once you've picked out three slots, you know that you must do pants/shirt/shoe in order in those three slots.

11:07 PM
Ohh it makes a little more sense now. So you can think of the slots as potential places for PantsX ShirtX and ShoesX , where x represents the customer (1 or 2 in our two person case)?
And whenever you are choosing 3 from the total slots, you are assigning them to PantsX, ShirtX, ShoesX to complete the outfit
And the combination formula handles all the possible arrangements

Right. And once you've grouped the 12 slots into collections of 3, we assign the first collection to person 1, the second to person 2, etc.
You need to stop and check whether we've overcounted, as that happens a lot in combinatorics/probability.

Makes a lot of sense.
Let me check if we overcount
Doesnt seem like we are

OK.

I'm curious. How would this problem change if each person could be given a random shoes, pants, and shirt. Regardless of what they ordered. Would we just end up multiplying by (4!)^3 since for each pants, shoes, and shirt from the collection of 3 chosen you will first have 4 options for the first person, then 3 for the next, then 2 for the next, then 1

So you only have one of each color and each person must take what he can get?

11:16 PM
Yea correct
4 distinct colors for the shoes, pants, and shirts

Yeah, I think you're right. You do the selection we already did, and then you fill in the choices for pants, etc., in order.

Ok cool. Thank you :)

Yea I didn't do the best job explaining it at first haha. Thanks for sticking through with me

There's a good lesson there for you — sometimes you want to group things together and count groups.
Like if you have a 8-letter word made out of 2 A's, 2 B's, 2 C's, and 2 D's. How many words are there?

11:20 PM
So here theres repetition
Like, i know the traditional way of doing it when you consider multisets
Would just be 8! / [(2!)(2!)(2!)(2!)]

OK, is this coming out the same as the way we just did your problem?

I don't think its exactly the same. Unless Im doing it wrong. Because wouldnt you be double counting with this example
since not everything is distinct

Do it exactly the way we did your pants/shirt/shoes thing.

Oh wow
(8C4)(4C4) = 280

Wait.
Where did 4's come from?

11:27 PM
Aren't we taking 8 total slots. And from there we are choosing 4 that will represent (A,B,C,D)
So at first we have 8C4
then when we've selected the first 4 slots we only have 4 remaining. So 4C4 to put the last four (A,B,C,D)

But you have lost track of what order the letters are going. This isn't a good approach.
We want actual "words."

hmm

Plus, you definitely have overcounting issues (and undercounting issues, as I already said). How do you proceed exactly as we did your problem?

What do you mean by "lost track of what order theyre going in"?

You only picked 4 slots. What order do the ABCD go in them?
So there are 4! words for each of your selections, but you've also overcounted.
I like the other way way better.

11:32 PM
I could see how it worked for our first example since P1 was distinct from P2, etc.

But we picked sets of 3 slots.

But here, I'm not sure how we could apply the same principle
Right, so why can't we pick sets of 4 slots here? and permute those

No, no, you're grouping the wrong things. Group the 2 A's into a group of 2. Group the 2 B's into a group of 2, etc.

If we pick sets of 3 slots, we'd still have to decide whether the sets would need (A,B,C) or (B,C,D) or (A,B,D) so 3 possibilities
Oh okay
Let me try that
Ah alright
So this eliminates any double counting since even if we have adjacent As, there is only 1 set of two items that will be adjacent in the same positions of the 8 slots
So it becomes (8C2)(6C2)(4C2)(2C2)

11:37 PM
Yea

OK, that's good. I'm done with you for now :)

One last extention question. So if we had 3 A's , 4 B's , 2 C's itd be the same approach but we are grouping into sets of whatever number of identical terms we are trying to arrange, correct?

Right.

So (9C3)(6C4)(2C2) for (arranging the A's)(arranging the B's)(arranging C's)
Okay cool :)
If only you were my professor haha

Yes, it gets more interesting if you can't have any adjacent pairs, etc.
I'm not a combinatorics person, but I did teach probability my last year before I retired. I had a lot of fun. If you try to learn multivariable analysis/calculus and linear algebra, you can "enjoy" my YouTube lectures. :)

11:45 PM
Where can I find your lectures?

They're linked in my profile on here.