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)
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?
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.
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: ruclips.net/video/Af3wHjdhZ6Y/видео.html. Note: this is not about whether this video is clear for smart enough people, but about clear teaching.
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
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, ...
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
Best explanation of this I've found online!
Awesome! Thanks for watching!
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?
Thanks for watching! I'll see what I can do about the finite selection video!
Must master the fundamentals! Thanks for the lesson.
Absolutely, competency in counting problems can be very powerful! Thanks for watching!
I finally understand it. Thanks!
Thanks that was really helpful.
Awesome, thanks for watching!
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)
Good explanation! thanks guy. Is there another course to explain how to count the number for fixed number of a1/a2/.../an?
Wow, that's just amazing! Many thanks! : )
My pleasure, glad you liked it!
Very wonderful
What about when selecting object in a group with twins and triplets which need not to be separated.
Thanks
is there an explanation for the problem at the end, where the infinite number of each element is changet to a fixed k number ?
वाह! क्या बात है।
Thank you Peter Griffin
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
What is the difference between permutation with repetition and combination with repetition?
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?
Awesome video, keep it up!
Thanks Daniel, I will! My next video will be a documentary, I hope you'll check it out!
Superb
thank you sir
Glad to help!
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.
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: ruclips.net/video/Af3wHjdhZ6Y/видео.html. Note: this is not about whether this video is clear for smart enough people, but about clear teaching.
Thanks 😊👍
My pleasure!
Thank you sir 🫡😭
Glad to help!
Why do we have n-1 bars didnt understand that idea
each bar represents one object except for a(n) because a(n) is the last one and no needed to stop.
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
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, ...
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
When order matter you should use permutations and combinations when order does not matter
Would suggest you to go through discrete maths by rosenreferrence book
brutal!! thnks 🤗
the video image is too poor, you need to fix it more