Hi everybody. I hope this helps clarify where the formula from my previous video comes from, if you have any questions or suggestions let me know. Thanks for watching 😀
@mariuszpopieluch7373 Жыл бұрын
Great and exhaustive explanation. Thank you. I have a question regarding this class of problems. Suppose we wish to generalise this approach and given a set of size n, consider all particular size partitions of all subsets of size k. The example would be, say using the technique in the video to select disjoint teams (of size i, j, k, say) from n people. Note that i+j+k=n. But suppose we want to see how many such teams we can create from m number of people, where m>n.
@VitalSine Жыл бұрын
Hello, Great question, and sorry for the delayed response. If I understand correctly, I believe this would be (m choose n) * (# of teams we can create from n people, using the technique in the video). Another way to write this would be (m choose i)(m - i choose j)(m - i - j choose k).
@hobouange582 Жыл бұрын
Thank you for your explanation. It was great. From Ivory Coast 🇨🇮
@VitalSine Жыл бұрын
I'm glad it was helpful!
@supreetshukla2 жыл бұрын
Best video on this topic. All clear. Thanks sir 🙏
@VitalSine2 жыл бұрын
Thank you, I'm glad you enjoyed the video 👍
@VitalSine Жыл бұрын
00:00 Stirling Number Review 03:00 Integer Partitions of Sets 04:30 Formula 05:27 Example 1 05:54 Example 2 07:00 Example 3 08:50 Formula Derivation 09:40 Intuition
@kevinleeds9794 жыл бұрын
i have studied it before so it is difficult to watch it through beginner eyes. there are many approaches to counting the partitions and i am waiting to see a recurrence equation, and a relationship to another combinatorial construction that I don't remember exactly, and also i think it has something to do with generating functions. i would like to go through the details myself on paper to review rather than see how another person has displayed it. this is probably my fault and not the video's fault! I don't think I have looked at this way of labeling the partitions, with the linear combination equaling n. The slides might show that the coefficients are nonnegative integers since negative coefficients would change the situation completely.
@VitalSine4 жыл бұрын
Thank you for watching and for the constructive feedback :)
@samreenfatima68733 жыл бұрын
Awesome explanation!!!!
@VitalSine3 жыл бұрын
Thank you!
@joaoafonso43253 жыл бұрын
Great video! Found it searching for a solution for a similar problem. I have similar containers, thus it simplifies to: n! / ( k! . (n/k)!^k), right? But my n elements can be similar, usually also in groups. Let's say n=12, with 4 letters appearing 3 times each. Any magic formula for these non-distinct n elements?