Counting Probabilities with Combinatorics and the Factorial

[@MrMaxusy]

You have a fantastic talent for teaching. Your videos on control theory helped me through my studies and now that I'm finished I still come back just to learn more about all sorts of topics, refreshers, whatever. Keep it up!

[@charlieawesome5986]

Thank you these videos are absolutely golden ☺.

[@ScienceMasterHK]

Waiting for a complete course Steve! Love it so far!

[@eliegakuba]

Thanks for these fantastic videos, Steve! Are there any accompanying exercises to help us grasp the concepts even better?

[@alarissacamila]

up up up!! i know we can probably just google it, but it would be nice to have some recommended list or any resources links

[@RolandoLopezNieto]

Great videos, thank you very much

[@alarissacamila]

I was so focused I got sad when the video ended. Is there any schedule for the next lessons? Im here for it either way, great explanation!!!

[@PrntScrpt]

Thanks professor

[@CaravaggioRoma]

thank you

[@magtazeum4071]

Lovely ❤

[@bitterbum1]

My attempt at the homework problem: "probability of no repeated letter" *PART 1: how many combinations exist fitting the description?* - no replacement - order does not matter (although letters are not repeated on the plate, unique letters can be in any order--not sequenced) - therefore equation to use is #3: n!/((n-r)!r!) n = 25 choices (letters of alphabet), r = 4 samples (letters on license plate) # letter combinations that fit description = 25!/(21!*4!) = *12650* *PART 2: how many possible license plate lettering combinations are there?* - replacement (each letter is found independently, with 25 possible choices each time) - order also does not matter (we are looking for letter groups, not letter sequences; a group is equivalent if they have the same letter contents) - therefore equation to use is #1: n^r / r! # possible lettering combinations = 25^4/4! = *16276* *PART 3: probability that a given sample out of all license plate lettering possibilities fits the description* P(A) = (# letter combinations that fit description)/(# possible lettering combinations) = 12650/390625 = *77.72%* *PART 4: Discussion* Do the numbers even affect this outcome? I feel like a version of the above equation could look like: P(A) = [probability letters fit description]*[probability numbers fit description] _rearranging so that descriptive terms are in numerator and all possible terms are in the denominator:_ P(A) = [# letter combinations that fit description * # number combinations that fit description]/[all possible letter combinations * all number combinations] Since all number combinations that fit the description of "no repeated letters" is equivalent to all possible number combinations in general, I would guess that those terms cancel out, and you are left with the original P(A) = 77.72% equation. I am not completely sure about this answer, though. Thanks so much, Steve! Stats was always my most difficult subject to understand on a conceptual level, mostly attributed to teachers and professors who refused to teach it in a conceptual manner.

[@igorg4129]

But what is the formula for a case "with replacment" when order "does NOT matter"?

[@studybuddy1015]

good video

[@AmirHX]

17:49 What is the probability that I don't have a repeated letter?---> 7.85% ((26)!/(26-4)!)+10^2 / 26^4+10^2

[@samsung6980]

I think it should not be added, but multiplied instead.

[@AmirHX]

@@samsung6980 the Numbers of Licenses?

[@Khazloo]

It should be multiplied. ((26)!/(26-4)!)*10^2 / (26^4)*10^2

[@dostdarali3999]

❤

[@benstallone6784]

n choices, r samples, without replacement, order matters: n! / (n-r)! n choices, r samples, without replacement, order doesn't matter: n! / ((n-r)!*r!) n choices, r samples, with replacement, order matters: n^r n choices, r samples, with replacement, order doesn't matter: ???????

[@benstallone6784]

never mind. found it n choices, r samples, with replacement, order doesn't matter: (n+r-1)! / (r!*(n-1)!) not obvious at all! you have to translate the problem to r objects being separated by n-1 separators

[@benstallone6784]

n! / ((n-r)!*r!) is always an integer?!? 🤨