I've been finishing up my discrete mathematics 1 course and next year I have discrete mathematics 2 so I can't wait to follow your very clear videos for help! Discrete maths in general is difficult for me to understand and I'm glad there are people like you out there that make it better for students that are really stuggling. Thank you for your videos Professor Brehm!
Glad I could help! I hope to have these videos done by the end of the summer, if not before, so they should be done in time for your new course. Hopefully there is some overlap in the topics.
I am not in any particular school at the moment. However, I would prefer your method of teaching over others any day. PS: I am also a waaaaay older coding student who is _starting_ to understand the high-level aspects of coding, thanks to you.
Nothing wrong with being WAAY older! Thanks for the comment! I waited a decade before getting my masters. It does make it harder to recall things you used to know, but more rewarding in the end.
Regarding the software company applicant problem, what if the set containing those 23 guys having other majors is not disjoint from the other two sets of majors? It's not stated explicitly that that set has no overlap with the other sets.
I'm decent in math and pretty good at finding shortcuts. So for the problem with 200 job applicants, my trick is: (107 + 116) - 200 = union... + 23 (the given # of applicants who didn't major in either) = 46. By the way, I'd have chosen a different sum so that there weren't two 23s, since some students will assume you can just double something. To make sure I was right, I tried it again with lower #s: 20 applicants. 12 majored in X, 11 in Y, and 4 in neither. Then you get... (12 + 11) - 20 + 4 = 7, which is the # in the center of the Venn diagram.
I have a question about the seating at a circular table. I hope I'm not splitting hairs or misreading the intent of the question, but it seems that there should only be half as many unique solutions (3 rather than 6). That's because as I interpret the question, it doesn't matter whether your friend A is on your right side or your left side. A shortcut to this problem is to think about who is sitting across from A; there are only 3 possibilities, so when A and B are across from each other, they can both hold hands with C and D (so to speak). In short, I'm guessing that -- at least as I interpret the question -- the answer is (n! ÷ n) ÷ 2 ... or just (n-1)! ÷ 2.
Oh btw, I need to study combinatorics. So, that other playlist is fine, right? What exactly is the difference between this one and that one? TIA Edit- two words
Hello Dear Mam, I don't have a premium subscription, and there are too many advertisements, sort of creates a distraction, but I can't afford the subscription pay.
I've been finishing up my discrete mathematics 1 course and next year I have discrete mathematics 2 so I can't wait to follow your very clear videos for help! Discrete maths in general is difficult for me to understand and I'm glad there are people like you out there that make it better for students that are really stuggling. Thank you for your videos Professor Brehm!
Glad I could help! I hope to have these videos done by the end of the summer, if not before, so they should be done in time for your new course. Hopefully there is some overlap in the topics.
your way of explaining is really easy to understand, so thank you!
The answer at 10:25 is 56 and not 46, for the software company problem
I am not in any particular school at the moment. However, I would prefer your method of teaching over others any day.
PS: I am also a waaaaay older coding student who is _starting_ to understand the high-level aspects of coding, thanks to you.
Nothing wrong with being WAAY older! Thanks for the comment! I waited a decade before getting my masters. It does make it harder to recall things you used to know, but more rewarding in the end.
thanks akkaya
You're just amazing, THANK YOU!!
Regarding the software company applicant problem, what if the set containing those 23 guys having other majors is not disjoint from the other two sets of majors? It's not stated explicitly that that set has no overlap with the other sets.
for some reason your teaching style is very effective for me I wish I could learn everything about life from you
I'm decent in math and pretty good at finding shortcuts. So for the problem with 200 job applicants, my trick is:
(107 + 116) - 200 = union... + 23 (the given # of applicants who didn't major in either) = 46. By the way, I'd have chosen a different sum so that there weren't two 23s, since some students will assume you can just double something. To make sure I was right, I tried it again with lower #s: 20 applicants. 12 majored in X, 11 in Y, and 4 in neither. Then you get...
(12 + 11) - 20 + 4 = 7, which is the # in the center of the Venn diagram.
I have a question about the seating at a circular table. I hope I'm not splitting hairs or misreading the intent of the question, but it seems that there should only be half as many unique solutions (3 rather than 6). That's because as I interpret the question, it doesn't matter whether your friend A is on your right side or your left side. A shortcut to this problem is to think about who is sitting across from A; there are only 3 possibilities, so when A and B are across from each other, they can both hold hands with C and D (so to speak). In short, I'm guessing that -- at least as I interpret the question -- the answer is (n! ÷ n) ÷ 2 ... or just (n-1)! ÷ 2.
there is video of bijective proof in combinatoric please? my university use this proof so much but they explained it bad its so hard
Oh btw, I need to study combinatorics. So, that other playlist is fine, right? What exactly is the difference between this one and that one?
TIA
Edit- two words
The other playlist uses a different textbook, but it is basically the same material. I'm redoing the course and using the Rosen text.
@@SawFinMath oh yes I compared the DM 2 playlist to Rosen and it's the same. I'll just study from that playlist.Thanks!
Hello Dear Mam, I don't have a premium subscription, and there are too many advertisements, sort of creates a distraction, but I can't afford the subscription pay.
I think make a math only account you won't get ads