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Combinatorics June 25 Chat
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AvatarGreg Miller
11:36
Closed and archived at 11:36  PM...
10:46
That's awesome... it's on "another level" of generating function problem!
Sabra
10:45
the first part made sense to me from the even options on the dice it was that second part with the odd numbers i really was not sure on but seeing the explanation now i get!
AvatarGreg Miller
10:41
if you wrap your brain around #8, then that's a big accomplishment.. that problem is in there basically as just a "here.. try this because its obnoxious" problem.. rather than a problem that is really meant to be a true "exercise".
Bethany
10:37
@Sabra you're welcome, glad I could help.
Sabra
10:34 
Thank you for the all the help #8 and #7 make more sense now.
AvatarGreg Miller
10:33
Looks like Bethany's comments on the "paint the vertices" #7 are good.. I just have to admit that I'm not a fan of that type of problem.  Then again, I'm not a geometry guy...
10:32
Not a fan of #7 on page 55... I'm fine with that as an omit... I probably shouldn't have assigned it.
10:31
That problem is a BEAST.. very difficult.  One of the hardest in the text overall.
10:30
Comment at 8:30... Problem 8 on Pg 54 is a beast.. probably the most difficult problem in the chapter.  The term in front of the bracket (in back of book) represents the three terms x2, x4 and x6 being unconstrained in any way.  Then ,the other three variables have constraints that are explained in the line right above the formula.  To model those constraints we pick one of x1, x3, or x5 to be even and the other two odd which gives us the term before the + sign... but it could also be (disjoint from what was just previously explained) that all three could be even, which gives us the term after the plus. (we have a + because of the "or")
Bethany
10:27
So we have four corners and for each corner we have the option of either black or white we can write this out one of two ways lets call black=z^o and white=z^1 so each corner has the generating function of z^0 +z^1. This gives us (z^0+z^1)^4=(1+z)^4 as our overall GF. Now we are looking for the coefficient on the z^2 term which is where we have the same number of white and black corners.
10:15
for Pg 55 #7 we are looking at the binomial theorem which I find confusing as as well but I will do my best to help out...
10:07
on pg 62 the reason that 1 is not in the answer is because you are only excluding that it appears exactly once so it could appear zero which is why the 1 is not included
Sabra
10:03
also on pg 62 #2 i got for my h(Z) = (e^z -1 - z)^5 however when i checked in the back of the book the answer was without the 1...did i make an error somewhere?
8:42
*drawing a blank
8:36
#7 on pg 55 i can draw it out to get the six ways but i am drawing a black on how to model that...each side only has two options (b or w)....so only thing i can think to say if 4 choose 2...which is 6 but i know it is not the model we are wanting
*an even
8:30
I am going back over some of the problems I struggled with the first time around. #8 on page 54. I understand the the way they wanted the equation set up and why...I also know evens added together make evens and two odds added together with make and even. My biggest issue is where to go from there....
AvatarGreg Miller
7:11
Chat is open.. actually been open, but I just got here to check at 7:12 PM...
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