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
>>16810837just count lol
>>16810837Less 4chan, more problems.https://calnewport.com/case-study-how-i-got-the-highest-grade-in-my-discrete-math-class/
>>16810838
>>16810837>over or undercountingWhat's an example of where you excluded or included too much?
>>16810841When 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
>>16810841Nvm, 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
>>16810844Here’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 MHow many if they have to be positive?I had never seen a problem like this before and had no idea how to approach it
>>16810854Glad 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?
>>16810859I 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
>>16810854Oh come on. Start with M=1,2,3,… and find the pattern. Read Larson’s problem solving book o algo.
>>16810866Appreciate 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”
>>16810861>we could have M and then n-1 many 0sYes. 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?
>>16810872I am so lost already lol. I might’ve already messed up but this is what I got
>>16810887Let's type>M = 1and>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 MWhat can you delete and what can you type?
>>16810896(1 choose 1)? Dude idk you’re already losing me
>>16810896Choose 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
>>16810897You 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 MHow 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?
>>16810901>We want n many things in Z+This makes it sound like you can't choose the same thing more than once.>>16810854This 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...
Let's go back to here>>16810844What's an example of where you excluded or included too much?
>>16810904Still not seeing it
>>16810918But what's an example of where you excluded or included too much?
>>16810918>I cannot tell when I am over or undercountingWhat's an example of this
>>16810914Even 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
>>16810924>>16810926>>16810922
>>16810926This is OP's tragedy.>I cannot tell when I am over or undercountingIf you're OP, tell me about the over or undercounting. Maybe I can lifeguard you to safety in my sexy pool boy speedo.
>>16810933Am down: will be your sexy damsel in distress if me putting on a wig and getting some implants is good enough iI described the issue here >>16810926it 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
>>16810837Sounds like you’re retarded anon. Sorry.
>>16810936Oh, that goes without saying. I’ve been a certified crayon eater since i was a child
>>16810935Don't play the lottery and never put on a wig or pantyhose.
>>16810841>my super secret strategy>study the fuck out of the material>shocked pikachu face.pngKys
>>16810953Shills get the rope. That means you.
>>16810854Pause: 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?!
>>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.
>>16810958:( ok mr smarty pants, how would YOU intend your students to find the solution?
>>16810854This 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.
>>16810962Start 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.
>>16810974I 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 :/
>>16810983I did this earlier in the thread. If M=1 there biggest sequence that adds up to 1 has length 1
>>16810990Ask your question again, then answer your question with your answer. Did your answer answer your question?
>>16810993Not 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 1,0,0If M=2 and n varies we get 1,1,0 If M=3 and n varies we get 1,2, 1I really am not seeing a pattern here
>>16811000Start 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.
>>16811003Are you eve reading my posts? I already did that>>16811000
>>16811013>Don't do this for every M, just for M = 1. Post your results for M = 1.
>>16811017>>16810918Did that earlier in the tread as well
>>16810953kys
>>16811018I'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.
>>16811028I 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
>>16811035>crtl f J3>1 resultDone.
>>16810837use a different definition for multiplication
>>16811112What did you have in mind anon?
>>16811143choose
>>16811146Idk what you mean. What are you looking for here?
>>16810854OP here. Did some more research. This is apparently a combinatorial question deeply related to something in physics called the Bose-Einstein problem>>16810974This 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)
>>16811157Then just move along, this thread is for people familiar with combinatorics.
>>16811160I literally made the thread. You cant tell me I don’t belong in my own thread
>>16811162You obviously don't if you don't even know about choose.
>>16811165Do 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
>>16811165>>16811169Sorry, 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?
>>16811172No, 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?
>>16811187Yes, 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
>>16811189I was saying to use choose in place of multiplication.
>>16811191Ah, that makes much more sense!Ok, can we approach the problem >>16810854We’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 >>16810974I 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
>>16811196Don't get too hung up on formalism, that attitude will hold you back
>>16810837Infinite 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 lolThis. 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.
Sea lion thread
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.
>>168108382+2=4
>>16811196I 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.
