Combinations with Repetition | Combinatorics

[@ophelialavey]

Best explanation of this I've found online!

[@WrathofMath]

Awesome! Thanks for watching!

[@Bedoroski]

An intuitive lecture that cleared up my mind. Can you make a video about the "finite objects selection" that was briefly mentioned at the end?

[@WrathofMath]

Thanks for watching! I'll see what I can do about the finite selection video!

[@punditgi]

Must master the fundamentals! Thanks for the lesson.

[@WrathofMath]

Absolutely, competency in counting problems can be very powerful! Thanks for watching!

[@MarcPhilipGoodman]

Everyone teaches the answer like this, but no one teaches the logic or idea of why and how someone up with the bars concept as a viable solution. So while I know how to answer it, I still wouldn’t be any smarter in understanding the process of coming up with the solution

[@So_aham]

you can see this problem like this - there are k (boxes to keep things) and n-1 sticks. So here we have to arrange k + n-1 objects that will be = (k+n-1)! / (n-1)!*k! ...(n-1 and k are identical objects so we are dividing by their factorial here) = (k+n-1) C (n-1) = (k+n-1) C (k)

[@孫傅康]

I finally understand it. Thanks!

[@elaborated-eggs]

I still do not understand why this approach works? How can I come to this conclusion on my own without being previously told the equation

[@MarcPhilipGoodman]

I know right! I don’t like being taught “this is how to solve it” I wanna understand how someone came up with the idea🧐

[@bartoszpominski9304]

Thanks that was really helpful.

[@WrathofMath]

Awesome, thanks for watching!

[@harrisondong5405]

Good explanation! thanks guy. Is there another course to explain how to count the number for fixed number of a1/a2/.../an?

[@MrDeadlyCrow]

Wow, that's just amazing! Many thanks! : )

[@WrathofMath]

My pleasure, glad you liked it!

[@charan_75]

What is the difference between permutation with repetition and combination with repetition?

[@WrathofMath]

Thanks for watching and the difference is that permutations are concerned with order, combinations are not. So consider selecting 4 letters, with repetition allowed. Then AABC and ABAC are two different permutations with repetition. However, they are the same combinations because they consist of the same letters. AABC has 2 As, 1 B and 1 C, and so does ABAC. Whereas ADFF and ADFG are different permutations and different combinations. Does that help?

[@Mu_utd]

Very wonderful What about when selecting object in a group with twins and triplets which need not to be separated. Thanks

[@adityaadit2004]

Thank you Peter Griffin

[@boeriucosmin4213]

is there an explanation for the problem at the end, where the infinite number of each element is changet to a fixed k number ?

[@bodaciouschad]

Wasn't this the process for *replacement* rather than repetition? Having infinitely many of each variable is just replacement. Repetition is allowing their to be a finite number, even those greater than 1, of a given "option" within the pool, i.e. [0,0,1,1,1,2,4] is repetition whereas your demonstration was predicated on replacement, i.e. the pool of options [1,2,3,4] each being replaced with each selection and having equal chances to be picked for future selections. They seem to be fundamentally different things.

[@harshkumar7686]

वाह! क्या बात है।

[@DanielSColao]

Awesome video, keep it up!

[@WrathofMath]

Thanks Daniel, I will! My next video will be a documentary, I hope you'll check it out!

[@nicchagall6075]

Superb

[@yingkaisong4258]

Not sure why some people said this was the best explanation. Clearly, it skipped explaining why it is picking n-1 out of k+n-1. The video has no indication that these "bars" may fit in the original k spots. It only shows that these bars may fit in between these k spots, and somehow you have k+n-1 in total. A better visualization would be draw k+n-1 spots and give examples to show how these bars can be in these k+n-1 spots. The following is definitely a better and convincing explanation: kzbin.info/www/bejne/d5eWqHugmc2NbLs. Note: this is not about whether this video is clear for smart enough people, but about clear teaching.

[@tangisiva4093]

thank you sir

[@WrathofMath]

Glad to help!

[@sziluv]

Thanks 😊👍

[@WrathofMath]

My pleasure!

[@krasimirronkov17]

Why do we have n-1 bars didnt understand that idea

[@harrisondong5405]

each bar represents one object except for a(n) because a(n) is the last one and no needed to stop.

[@guy_you_can_trust]

imagine multiple compartments for storing money and k copies of each bill, theres only enough space to accommodate having 3 bills per compartment - the remaining 'room' in a compartment can accommodate k-1 bills after choosing a bill i.e. if n=4 and k bills = 3 (imagine those bills being $1, $5 and $20) then placing a $1 bill in its 'corresponding' box will leave k-1 bills of space left (think vertical height from where it's placed) this means we initially have n+k-1 'spaces' to work with

[@eliotvarda]

He said, WE COUNT BARS, that n-1 originates not from a(n), but from first element a(1), if your combination is just placing a(1) on every place, than you don't need any bars, for that reason, i think, ...

[@HoudaImeneMECHKAK]

Thank you sir 🫡😭

[@WrathofMath]

Glad to help!

[@RelixAMV]

Ok so we know how to work permutation and combinations but how do we distinguish which we're supposed to find from just the question? Cuz I could not tell the difference between some of these questions with the ones u did on permutation

[@jadejaprem5692]

When order matter you should use permutations and combinations when order does not matter

[@jadejaprem5692]

Would suggest you to go through discrete maths by rosenreferrence book

[@omarluiscamposnaranjo953]

brutal!! thnks 🤗

[@HongBui-rf2te]

the video image is too poor, you need to fix it more