Discrete Math II - 6.3.3 More Combinations and Combinatorial Proof

[@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 :)

[@navaneethsankars]

The best Discrete Maths class ever!😌

[@SewonKimMusic]

Brilliant tutorial. Wish universities had professors like you!

[@SawFinMath]

I am a university professor ❤️

[@calebelizon5810]

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

[@melhzt5024]

You helped me a tone

[@netanelkaye3014]

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..

[@SawFinMath]

Of course we are. We have slots of heads and tails and are choosing how many go on each.

[@pizzaablaze]

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

[@deuce2293]

I got A thanks to your videos, thanks!

[@bobdobson3800]

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.

[@LucasFe02]

I see you tried to use inclusion-exclusion principle to solve this problem, which is a valid solution. The overall path we need to follow is (i) find all the permutations of the word ADAMANT, (ii) find all the anagrams with three consecutive A's, (iii) find all the anagrams with two consecutive A's and (iv) subtract the anagrams from all the permutations. (i) Pemutation of ADAMANT: 7!/3!=840 / Since there are no differences between the A's we need to divide by 3! or else we would be counting the permutations of them. (A1)D(A2)M(A3)NT = (A2)D(A3)M(A1)NT and so on... (ii) Anagrams with three consecutive A's: |AAA|DMNT --> 5! = 120 (iii) Anagrams with two consecutive A's: _ |AA|_D_M_N_T _ , the last A has 6 - 2 = 4 places to be, hence he cannot stay at the left or the right of the block of the A's because we would be overcounting what we've done in step (ii). 5!*4 = 480 (iv) 840 - 480 - 120 = 240 Please let me know if you didn't understand something or I made a mistake.

[@ThomasJohnson-xp8ue]

@@LucasFe02 Just want to say thanks for providing such a good response! I made a similar mistake in my first try at solving this problem (mistakingly counting the choice of A1 A2 and A3 as unique outcomes), and your response was incredibly helpful.

[@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 HHHTTTTT = TTTTTHHH)? 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!

[@melhzt5024]

God bless your soul

[@pseudovictim]

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

[@SawFinMath]

I've actually never heard of the stars and bars method. So maybe?

[@danielsarmiento7545]

i don't understand:(