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Combinatorics June 3 Discussion
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AvatarGreg Miller
Good night.. all..
10:03
Keep working hard.. allow yourself to move back and forth between the nagging parts of problems.. we are in the key learning period for Chapter 1.. how many problems you work between now and Friday night will determine your understanding...
10:02
Gotta close this down about a half hour earlier than normal today... Good to pop in here from time to time and see several of you chatting to each other...
10:01
Tomorrow's videos:  ALL EXAMPLES.  Not moving forward in Martin.  We need to time to work ( A LOT) of these problems tomorrow.. next five or six videos are all just examples of various problems and usages of techniques...
10:00
You're welcome.
Bethany
9:55
Thank you sir. That helps a lot as I really thought I had that right.
AvatarGreg Miller
9:53
Bethany.. FYI.. the answer in the back of the book for #6 on page 15 has a typo.  it should read 11 choose 8 with repetition (angled brackets).. not 11C8... repetition is allowed there... Martin error.
Bethany
9:52
I see now, I was misreading and thought there were only 4 people total.
AvatarGreg Miller
9:50
But, to motivate:  There 4n people in the problem, 4 from each country.. n total countries.. but, we want to "pair up" people from the same country... so we have to first ask.. if we have four people from the USA.. how many arrangements can we create of the two groups of pairs... see if you can reason that the answer to that part of the problem is 12.
Bethany
9:50
Anyone understanding why #6 on pg 15 doesn't have repetition allowed in the wedge selection for vowel placement?
AvatarGreg Miller
9:48
Bethany... I have chosen the "nationalities" question to cover in my "Examples Videos" series tomorrow...
Bethany
9:42
On page 15 #10 I am completely lost on the nationalities. Can a person have more than one nationality? Why is the book using place cards and how is the 'n' nationalities what we are choosing as r?
AvatarGreg Miller
9:15
So, our answer is 26C10 - 17C10.  This problem is quite unique.. and is surely one of the most challenging one's we've faced so far.  But, there are a lot of very tough problems right through here...
9:14
Why does this work?  Because when we are done, we will have a string of 26 characters... all R's and C's.. that meet the condition that no two C's are adjacent.  We can then "lay that string" of 26 characters over the top of the alphabet and it will tell us which letters to pick.
9:13
#11 on Page 9 requires a very interesting way of thinking.  The problem as directly stated is too difficult to deal with.. so instead we'll count the number of 10-letter subsets that DO NOT have consecutive letters because the phrase "has a pair of consecutive letters" and "DO NOT have consecutive letters" are disjoint and opposite of one another.  So, the answer is 26C10 minus the answer to our NEW problem.  Now, toward this new problem.  10 of the letters will be chosen(C) and 16 of the letters will be rejected(16).  It's the chosen one's that have the restriction of now NOT being adjacent.  So, imagine placing down 16 R's (1 way to do it).  These 16 R's form 17 wedges on which the C's could be placed.  Of these 17 wedges, we want to pick 10 of them:  17C10 ways.  Then, we can place the C's one the chosen wedges in 1 way.  So, this leads to a total count of 17C10 for the new problem.
David
9:05
So nothing between the two pluses. Thanks.
Dayna
8:26
Is page 9, #11 the 17 choose 10 is similar to the fence posts 2 yards apart, total of 20 yards from the notes? I am struggling with this one too.
Bethany
8:13
The pluses each repesent a 'D' or divider and the 1 are the 'U' or units so UDDUUUDUUU is 1++111+111 is 1+0+3+3=7. I try to think of the pluses as the dividers/spaces between the boxes where you can place units/objects.
David
7:36
Why are two pluses a zero but one plus is just a space?
Bethany
6:12
On page 15 #5 and 6 I understand no repetition allowed in wedge selection on #5, but on #6 shouldn't you be able to put two or more vowels in the same slot if desired? It did not specify that the vowles could not be adjacent.
5:31
I am struggling to understand # 11 on page 9. How to select subsets with a pair of consecutive letters. The back of the book method of subtracting out the subset with non-consecutive letters is not helpful as I do not see how they are adding 1 to 16 and choosing 10.
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