Discrete Math II - 6.3.3 More Combinations and Combinatorial Proof

Just a few practice questions involving permutations and combinations, along with a brief introduction of combinatorial proof. Don't worry! You will see this again and again!

Video Chapters:
Intro 0:00
The First Two Practice 0:15
Another Practice and an Identity 5:03
Combinatorial Proof 7:18
One Last (3-part) Practice 8:35
Up Next 11:39

This playlist uses Discrete Mathematics and Its Applications, Rosen 8e

Power Point slide decks to accompany the videos can be found here:

The entire playlist can be found here:

[melodyjadyn]

In case this confuses anyone, at 11:44, the fraction is meant to read 52!/(47!5!). She says 52 here, but wrote 42 by mistake. Great videos here, they're super helpful :)

[SewonKimMusic]

Brilliant tutorial. Wish universities had professors like you!

[bobdobson]

Can someone help find where is the mistake in doing this for the "adamant" question? I solve the complement of the problem and find the total number of words with 2 consecutive A's plus the total number of words with 3 consecutive A's. Since we can treat the grouped A's as one letter, this is 6! and 5! respectively. Then minus that from the total number of words to get the number of words without consecutive A's (7! - 6! - 5!) = 4200

My answer is very wrong but I can't see the logical flaw in my argument, any help is appreciated.

[calebelizon]

11:15 why can't I just subtract 1 from k to account for the case that they are all men and do 13 choose 9? The answer ends up being way too high and I don't understand why at all

[deuce]

I got A thanks to your videos, thanks!

[pizzamazin]

Hello Kimberly ty for your videos they helped me a lot and I got an A thanks to watching your videos!

[Mkpjetra_ncsu]

I am a bit confused on the first coin flip question ~3:05, it seems like it should be a combination problem but the way it is analyzed seems more like a permutation. If the order of the coins doesn't matter, then wouldn't counting 2^8 possibilities introduce redundancy (ex: when =

I tried counting the total options it as if it is a combination with 8 objects and 2 distinct states (H or T), which gave me (8+(2-1))/(2-1) = 9 possible combinations, which made sense as it could be all Heads (state 1), all Tails (state #2), or any number of Heads 1-7 (states #3-9), where the number of heads and tails is complementary. I then subtracted the 3 undesirable states (0H, 1H, 2H), yielding 6 possible combinations.

Let me know if I am looking at this wrong!

[netanelkaye]

How exactly is the coin flip example at around 3:00 a N choose K problem? We're not choosing to put N things into K slots..

[pseudovictim]

Is the first problem (the word "adamant") also called "stars and bars" method?