Combinatorics of Set Partitions [Discrete Mathematics]

[@VitalSine]

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]

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]

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]

Thank you for your explanation. It was great. From Ivory Coast 🇨🇮

[@VitalSine]

I'm glad it was helpful!

[@supreetshukla]

Best video on this topic. All clear. Thanks sir 🙏

[@VitalSine]

Thank you, I'm glad you enjoyed the video 👍

[@VitalSine]

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

[@kevinleeds979]

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.

[@VitalSine]

Thank you for watching and for the constructive feedback :)

[@samreenfatima6873]

Awesome explanation!!!!

[@VitalSine]

Thank you!

[@joaoafonso4325]

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?

[@queenlowkey9481]

I hate this whole combinatorics topic🚮