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Thread 16810837

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Anonymous No.16810837 [Report] >>16810838 >>16810841 >>16810844 >>16810936 >>16811112 >>16811303 >>16819005 >>16823419
How do I make combinatorics make sense?
I’m really struggling with this stuff. I just bombed my exam and I really need to learn this stuff. I know the best solution for math is just to do more problems, but I feel like that only helps if you’re gaining an intuition behind what the heck you’re doing. I’m so stumped on these that I have to use chat GPT to help me even know where to start. I cannot tell when I am over or undercounting and while once I see the answer and the explanation I can see how they got there I often times would have no idea how I’d arrive there on my own.

I’m not really seeing a pattern here and I know I’m not stupid, I’ve taken some advanced courses before that I’ve done well in and those have been tough too, but especially with the proof based ones if you looked at the definitions and relevant results you could sorta get it by smashing them together for long enough.

Here I’m just given a problem and expected to pull the answer out of thin air.

I’m stating to think algebraists are the mathematicians with all the talent, and that people who like analysis like me are just monkeys and typewriters


Please help lol
Anonymous No.16810838 [Report] >>16810842 >>16811303 >>16811537
>>16810837 (OP)
just count lol
Anonymous No.16810841 [Report] >>16810849 >>16810851 >>16810953
>>16810837 (OP)
Less 4chan, more problems.
https://calnewport.com/case-study-how-i-got-the-highest-grade-in-my-discrete-math-class/
Anonymous No.16810842 [Report] >>16822565
>>16810838
Anonymous No.16810844 [Report] >>16810854 >>16810914
>>16810837 (OP)
>over or undercounting
What's an example of where you excluded or included too much?
Anonymous No.16810849 [Report]
>>16810841
When does it end? I get I gotta do more, but I did all the problems in all the HWs, did all the review problems AND had ChatGPT make up practice problems for me. I’m just not getting it man. There has to be a better way
Anonymous No.16810851 [Report]
>>16810841
Nvm, read the thing. That is the most braindead way to get good grades possible. Sadly it is also effective. Guess I have more problems to do
Anonymous No.16810854 [Report] >>16810859 >>16810866 >>16810896 >>16810904 >>16810912 >>16810956 >>16810974 >>16811158 >>16811196 >>16819023 >>16819046
>>16810844
Here’s a problem from the exam I bombed (I’m changing a couple things in case my professor somehow visits /here/)

How many finite sequences of n many non negative integers are there that add up to some natural number M

How many if they have to be positive?

I had never seen a problem like this before and had no idea how to approach it
Anonymous No.16810859 [Report] >>16810861
>>16810854
Glad to see your prof starts the natural numbers from 1. You're in good hands.

So there are two questions. One about non negative and one about positive. What makes those different questions?
Anonymous No.16810861 [Report] >>16810872
>>16810859
I get we can have 0 in some and not in the others. That part I can wrap my head around. I don’t see what the count would be for general form of either though.
Like I get we could have M and then n-1 many 0s and there are n! Ways to place that M, right? But I couldn’t get the general form
Anonymous No.16810866 [Report] >>16810869
>>16810854
Oh come on. Start with M=1,2,3,… and find the pattern. Read Larson’s problem solving book o algo.
Anonymous No.16810869 [Report]
>>16810866
Appreciate the book recommendation anon. I will spend half an hour trying to intuit a pattern but I do not have high hopes for being able to just “get it”
Anonymous No.16810872 [Report] >>16810887
>>16810861
>we could have M and then n-1 many 0s
Yes. Don't worry about placing them in order yet.

Then start from the easiest thing you can write out. M = 1. What happens to the two questions when n = 1? n = 2? n = 3?
Anonymous No.16810887 [Report] >>16810896
>>16810872
I am so lost already lol. I might’ve already messed up but this is what I got
Anonymous No.16810896 [Report] >>16810897 >>16810901
>>16810887
Let's type
>M = 1
and
>n many = 1.
This means you can delete M and type 1. You can delete n many and type 1.

Let's see words from (>>16810854)
>How many finite sequences of n many non negative integers are there that add up to some natural number M
What can you delete and what can you type?
Anonymous No.16810897 [Report] >>16810904
>>16810896
(1 choose 1)? Dude idk you’re already losing me
Anonymous No.16810901 [Report] >>16810912
>>16810896
Choose an in Z+ and M in N. We want n many things in Z+ s.t. Sum from i=1 to n of ai =M
Anonymous No.16810904 [Report] >>16810918
>>16810897
You already know that 0 isn't positive. The next step is to see what happens when M = 1.

>>16810854
>1) How many finite sequences of n many non negative integers are there that add up to some natural number M

How many finite sequences of 1 non negative integers are there that add up to some natural number 1?
How many finite sequences of 2 non negative integers are there that add up to some natural number 1?
How many finite sequences of 3 non negative integers are there that add up to some natural number 1?

>2) How many if they have to be positive?

How many finite sequences of 1 positive integers are there that add up to some natural number 1?
How many finite sequences of 2 positive integers are there that add up to some natural number 1?
How many finite sequences of 3 positive integers are there that add up to some natural number 1?
Anonymous No.16810912 [Report]
>>16810901
>We want n many things in Z+
This makes it sound like you can't choose the same thing more than once.
>>16810854
This makes it sound like you can.
The answers would be different but the best thing is always to start at M =1 and see what happens as n = 1, 2, 3...
Anonymous No.16810914 [Report] >>16810926
Let's go back to here
>>16810844
What's an example of where you excluded or included too much?
Anonymous No.16810918 [Report] >>16810922 >>16810924 >>16811018
>>16810904
Still not seeing it
Anonymous No.16810922 [Report] >>16810927
>>16810918
But what's an example of where you excluded or included too much?
Anonymous No.16810924 [Report] >>16810927
>>16810918
>I cannot tell when I am over or undercounting
What's an example of this
Anonymous No.16810926 [Report] >>16810927 >>16810933 >>16810935
>>16810914
Even simple stuff like doing poker hand calculations. If we’re doing likelyhood of the hand being 2 from the same suit, 2 from a different suit and then one leftover i thought that would be (13 choose 2) but then there are 4 suits that can be and you’re choosing one of them so that’s (4 choose 1) and then you have to choose 2 cards from another suit (13 choose 2) but now it has to be a different suit so it’s now (3 choose 1) and then we have to multiply by the choices for the last card (26 choose 1) and divide by all possible choices (52 choose 5)

That was what I wrote and that made sense to me, but oops! That’s completely wrong. Instead of being (4 choose 1) times (3 choose 1) it’s (4 choose 2) but that doesn’t make sense to me at all
Anonymous No.16810927 [Report]
>>16810924
>>16810926
>>16810922
Anonymous No.16810933 [Report] >>16810935
>>16810926
This is OP's tragedy.
>I cannot tell when I am over or undercounting
If you're OP, tell me about the over or undercounting. Maybe I can lifeguard you to safety in my sexy pool boy speedo.
Anonymous No.16810935 [Report] >>16810952
>>16810933
Am down: will be your sexy damsel in distress if me putting on a wig and getting some implants is good enough i

I described the issue here >>16810926
it can’t get more blatant than that desu.

I don’t really have a feel for what the answer SHOULD be, but i was actually kinda confident i got that one right till I doubled checked after.


Let’s break that down specifically. Why is that 4 choose 2 as opposed to 4choose 1 times 3 choose 1?

The second suit can’t be one of the first right? I guess i can say that we’re picking 2 of the 4 suits for each of them but you could’ve told me either answer was correct and I’d be able to convince myself that made sense. No real feel for what kinda thing I should get here
Anonymous No.16810936 [Report] >>16810937
>>16810837 (OP)
Sounds like you’re retarded anon. Sorry.
Anonymous No.16810937 [Report]
>>16810936
Oh, that goes without saying. I’ve been a certified crayon eater since i was a child
Anonymous No.16810952 [Report]
>>16810935
Don't play the lottery and never put on a wig or pantyhose.
Anonymous No.16810953 [Report] >>16810954 >>16811025
>>16810841
>my super secret strategy
>study the fuck out of the material
>shocked pikachu face.png
Kys
Anonymous No.16810954 [Report]
>>16810953
Shills get the rope. That means you.
Anonymous No.16810956 [Report] >>16810958
>>16810854
Pause: asked gpt to help explain and it says I can see this in terms of the sequence being ~NegativeBinomialy?

I never would’ve thought of that but that would erase the voodoo magic of combinatorics.

The stuff actually cancels out beautifully. Would my professor make that the intended solution? Is that too big brain?!
Anonymous No.16810958 [Report] >>16810962
>>16810956
>Freeze frame
>I found the bible code in a LLM
>Would my professor make that the intended solution?
If he did, you should immediately focus on making friends with rich classmates and also transferring to a better school.
Anonymous No.16810962 [Report] >>16810983
>>16810958
:( ok mr smarty pants, how would YOU intend your students to find the solution?
Anonymous No.16810974 [Report] >>16810989 >>16811158 >>16811196
>>16810854
This sounds like a weak composition (stars and bars) problem if I'm reading it right. Look that up, its way simpler than everyone here is making it seem.

The gist is that if we want an (ordered) sequence of [math]n[/math] nonnegative integers which add up to [math]M[/math], we can think of it as putting [math]M[/math] "ones" into the parentheses in the expression:

[math] \underbrace{(\quad) + (\quad) + \dotsb + (\quad)}_\text{$n$ times} = M[/math]

The key is that we can represent this in a simple way. Imagine using [math]\star[/math] to represent a "one". We have [math]M[/math] total [math]\star[/math]'s that we need to divide among [math]n[/math] terms. We could do this by putting all [math]M[/math] [math]\star[/math]'s in a row, then using [math]n-1[/math] "dividers" to break them into [math]n[/math] terms.

For example: To represent [math]3 + 2 + 0 + 1 = 6[/math], we could draw:

[math]\star \star \star \mid \star \star \mid \quad \mid \star[/math]

Think about how many different orderings there are of [math]M[/math] [math]\star[/math]'s and [math] n-1[/math] [math]\mid[/math]'s. Also, think about how you could (slightly) modify the setup if we insist that the integers must be positive.
Anonymous No.16810983 [Report] >>16810990
>>16810962
Start by writing out how many sequences of length n add up to M = 1, starting from n = 1, 2, 3..., if you're not allowed to use a zero. This how it normally works. Then write out what happens if you are allowed to use a zero.

Don't do this for every M, just for M = 1. Post your results for M = 1.
Anonymous No.16810989 [Report]
>>16810974
I think we might’ve touched on that briefly in class but that doesn’t seem like something that’s super easy to see.

In fact, I know I did because I went to office hours and asked about that exact thing because I had trouble understanding it.

Guess I only have myself to blame then :/
Anonymous No.16810990 [Report] >>16810993
>>16810983
I did this earlier in the thread. If M=1 there biggest sequence that adds up to 1 has length 1
Anonymous No.16810993 [Report] >>16811000
>>16810990
Ask your question again, then answer your question with your answer. Did your answer answer your question?
Anonymous No.16811000 [Report] >>16811003 >>16811013
>>16810993
Not in the slightest. I wrote out the terms for n from 1 to 3 and M from 1 to 3
If M=1 and n varies we get 1,0,0
If M=2 and n varies we get 1,1,0
If M=3 and n varies we get 1,2, 1

I really am not seeing a pattern here
Anonymous No.16811003 [Report] >>16811013
>>16811000
Start by writing out how many sequences of length n add up to M = 1, starting from n = 1, 2, 3..., if you're not allowed to use a zero. This how it normally works. Then write out what happens if you are allowed to use a zero.

Don't do this for every M, just for M = 1. Post your results for M = 1.
Anonymous No.16811013 [Report] >>16811017
>>16811003
Are you eve reading my posts? I already did that>>16811000
Anonymous No.16811017 [Report] >>16811018
>>16811013
>Don't do this for every M, just for M = 1. Post your results for M = 1.
Anonymous No.16811018 [Report] >>16811028
>>16811017
>>16810918
Did that earlier in the tread as well
Anonymous No.16811025 [Report]
>>16810953
kys
Anonymous No.16811028 [Report] >>16811035
>>16811018
I'm on a laptop and can read sideways things by turning it sideways. You didn't do what you said you did. Last chance to be sincere and post your results for M = 1.
Anonymous No.16811035 [Report] >>16811040
>>16811028
I am going to crash the fuck out.
If M, the natural number we are saying the sequence has to add up to, is 1, and n is in J3 we get 1 possible configuration for n=1, m=1 {1}, 0 possibilities for the rest since the smallest natural is 1 so you can’t have a sequence of more than 1 natural term that adds to 1.

Please don’t gaslight me man, I’m really trying here
Anonymous No.16811040 [Report]
>>16811035
>crtl f J3
>1 result
Done.
Anonymous No.16811112 [Report] >>16811143
>>16810837 (OP)
use a different definition for multiplication
Anonymous No.16811143 [Report] >>16811146
>>16811112
What did you have in mind anon?
Anonymous No.16811146 [Report] >>16811157
>>16811143
choose
Anonymous No.16811157 [Report] >>16811160
>>16811146
Idk what you mean. What are you looking for here?
Anonymous No.16811158 [Report] >>16819027
>>16810854
OP here. Did some more research. This is apparently a combinatorial question deeply related to something in physics called the Bose-Einstein problem

>>16810974
This anon gave the only explanation that’s made any sense to me so far, but even then I’m not all the way there (really just leading me to disliking combinatorics more)
Anonymous No.16811160 [Report] >>16811162
>>16811157
Then just move along, this thread is for people familiar with combinatorics.
Anonymous No.16811162 [Report] >>16811165
>>16811160
I literally made the thread. You cant tell me I don’t belong in my own thread
Anonymous No.16811165 [Report] >>16811169 >>16811172
>>16811162
You obviously don't if you don't even know about choose.
Anonymous No.16811169 [Report] >>16811172 >>16819790
>>16811165
Do you get off on feeling superior about having this esoteric knowledge that is barely applicable and then shaming others who don’t get it as easily? You would be a terrible teacher
Anonymous No.16811172 [Report] >>16811187
>>16811165
>>16811169
Sorry, I was getting a little heated.

I don’t quite see what you’re asking though. Do you mean choosing with/without order mattering? With/without replacement?
Anonymous No.16811187 [Report] >>16811189
>>16811172
No, I am referencing choose as it is used in combinatorics maybe you know it as CHOOSE and are just to dense to make the connection?
Anonymous No.16811189 [Report] >>16811191
>>16811187
Yes, insulting me doesn’t help, but now I see you meant (nk) I assumed you meant choose a different method of multiplication and was confused
Anonymous No.16811191 [Report] >>16811196
>>16811189
I was saying to use choose in place of multiplication.
Anonymous No.16811196 [Report] >>16811259 >>16811572 >>16819029
>>16811191
Ah, that makes much more sense!

Ok, can we approach the problem >>16810854

We’ve established that it is a composition problem, what an application of it is, that when we include 0 it’s called a weak composition problem, and we have been given a method of approach by >>16810974
I have a lot easier of a time working in mathematical formalism and using proof.

Is there a chance you could give me a rigorous write up for this?

The “stars and bars” feels so wishy washy I can’t believe it even works
Anonymous No.16811259 [Report]
>>16811196
Don't get too hung up on formalism, that attitude will hold you back
Anonymous No.16811303 [Report]
>>16810837 (OP)
Infinite dimensional hyper-urns containing one or more red, black, and bluwn socks.
You pick one card.
Look at the card. LOOK AT ALL THOSE GLOWING URNS AND TOKENS!
Return to card.
>>16810838
>just count lol
This. But honestly, I could never remember when to start with 1 or when to start with 0. I'd take an average and let it be ə > 0.
Anonymous No.16811319 [Report] >>16814058
Sea lion thread
Anonymous No.16811340 [Report]
I feel you zoomer. Combinatorics and probability and statistics are really weird things. For each problem you have an ad-hoc way of solving. I absolutely hated that. But try a few solved example problems on your own and you can slowly find a way.
Anonymous No.16811537 [Report]
>>16810838
2+2=4
Anonymous No.16811572 [Report]
>>16811196
I would partially disagree with anon on formalism. I think it's good to have an instinct for rigor, you just have to make sure you're willing to accept "it is what it is" when the fully rigorous explanation is beyond you at the moment.

As regards your question: The way you prove simple combinatorial arguments like this work is to establish a bijection between the two things you are counting, in this case weak compositions and "stars and bars" pictures. Once you establish a bijection between two sets (and thus prove they have the same cardinality), you're fully justified in counting whichever is easier.

As you'll recall, to prove bijectivity you need to show:
(1) Injectivity. (i.e., show that there aren't two compositions with the same stars and bars representation)
(2) Surjectivity. (i.e., show that every valid stars and bars picture corresponds to a composition.)

I would definitely suggest you focus on figuring out how to count the stars and bars first (I promise it's not that hard), but once you do that it's not too difficult of an exercise to prove the bijection if you want to be completely rigorous.
Anonymous No.16814058 [Report]
>>16811319
kek
Anonymous No.16819005 [Report]
>>16810837 (OP)
You'll be alright
Anonymous No.16819023 [Report]
>>16810854
stars and bars was my most immediate thought. Looks like others mentioned it
Anonymous No.16819027 [Report]
>>16811158
>deeply related
I mean, the word deeply isn't my favorite word here. That's like saying the use of sugar is deeply related to making cakes. Like, okay? Was deeply needed?

In QM - and this is just undergrad stuff - there's this funny story of how the math at the time wasn't adding up concerning bosons and fermion particles, in that it wasn't predicting the right things. Bose sends a paper to Einstein about how the math he does is correct, and Einstein quickly understands that Bose is interpreting the particles in a completely new way before. When Bose comes to visit Einstein in the future, he is asked if he knew what he had done, and it seems that for the rest of his life, in telling this story, Bose always was forward in stating he did not.

It seems that in many ways, particles can be indistinguishable from each other. All electrons look like electrons, all protons look like protons, etc. If you have one electron here, and another a mile away, of course you can distinguish them by position, but what happens when they're close to each other? Their waves interact in a non-neglible way so it's hard to distinguish.

For your stars and bars thing, notice that all the stars look the same. It's the same deal.

>even then I’m not all the way there
What's your issue?

Stars and bars is something you just keep in the back of your head. There are problems out there where you don't even realize it's a stars and bars problem, but it's just that there's a way to rearrange how you think about the problem to make it one. Doing practice problems really is the best way to get into it, like you saw in the other comments.
Anonymous No.16819029 [Report] >>16819702
>>16811196
Not him, what do you mean by rigourous? What part do you not understand from the other guy's answer? Stars and bars is pretty straightforward imo
Anonymous No.16819046 [Report]
>>16810854
Still no solution.

Grim.

Can we all just stop larping? I personally dropped out from my program and shit at math. I assume all of you guys are similar.
Anonymous No.16819702 [Report] >>16819712 >>16819722
>>16819029
I was looking for a formal proof. That is not rigorous whatsoever. That is a heuristic
Anonymous No.16819712 [Report]
>>16819702
I mean, it's a simple math trick. Did you understand what the guy was trying to convey? Like, can you solve the problem you originally posed using stars and bars? Cause if you already understand it (you said "Ah that makes sense" before), the proof that it works seems pretty obvious.

If you can't already solve the problem, what part of the stars and bars do you not get?

Like, I learned stars and bars in HS, this wasn't meant to be a hardcore problem.
Anonymous No.16819722 [Report]
>>16819702
You said you struggled with combinatorics, so I'm not certain if you know how to answer the following:

When someone says something like 7 Choose 2, what is the cliche reason someone would talk that number. Like, can you give me an example problem where 7 Choose 2 is the answer? The symbol for this would be something like [math]
\left( \begin{gathered} 7 \\ 2 \end{gathered} \right)
[/math] . Maybe you can abuse this to say this is the same as [math]
\left( \begin{gathered} 7 \\ 2, 5 \end{gathered} \right)
[/math] .

Would you understand what I mean if I mention something like [math]
\left( \begin{gathered} 7 \\ 2, 2, 3 \end{gathered} \right)
[/math] , and what sort of example problem would lead to an answer like this?
Anonymous No.16819790 [Report]
>>16811169
It's extremely applicable. As someone said before, you could use it to predict poker cards in a deck. This would certainly increase the accuracy of your decision making whilst gambling. It certainly has for me
Anonymous No.16819959 [Report]
I like combinatorics but am really bad at them. A lot of practice helped, but I still feel like I am missing some fundamental information.
Anonymous No.16822565 [Report]
>>16810842
thanks tony
Anonymous No.16823419 [Report]
>>16810837 (OP)
Don't try to understand factorials or the binomial coefficient (choose) formula.
Just memorize the formulas.
t. 135 IQ. I intuitively understand the idea behind factorials, but I find it futile trying to get an intuitive understanding behind binomial coefficient.