Discrete Math II - 6.1.3 The Subtraction and Division Rules

We finish up section 6.1 by discussing the last two basic counting rules; the subtraction and division rules. The subtraction rule is really just the principle of inclusion-exclusion. You might also see it called the General Addition Rule, as it is about "or" questions and addition of possible outcomes. The "difference" (sorry, dumb math humor) is that we subtract any elements that were counted in both sets we added.

The division rule isn't one we use often, but we go through one quick example to show how to account for options that were counted multiple times.

Video Chapters:
Intro 0:00
The Subtraction Rule Example 0:15
The Subtraction Rule Formalized 5:59
The Division Rule 10:41
The Division Rule Example 11:32
Up Next 13:49

[DivinityAgarioMore]

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!

[gyfennn]

your way of explaining is really easy to understand, so thank you!

[KrrithikEzhilarasan]

The answer at 10:25 is 56 and not 46, for the software company problem

[egok]

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.

[davidstone]

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.

[GraceandWisdom]

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[ea]

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[davidstone]

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.

[techjesus]

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[marik]

there is video of bijective proof in combinatoric please? my university use this proof so much but they explained it bad its so hard

[mango-strawberry]

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

[aayushsugandh]

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