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Combinatorics June 29 Chat
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AvatarGreg Miller
10:58
Closed and archived at 10:58 PM...
10:10
Sabra.. for #1.. try actually drawing pictures for the n=2 and n=3 case and see if you can match up those specific cases with what Martin is discussing in the "Back"
10:07
I'll have to circle back around to the follow up questions on #5 that was previously discussed.  I don't have my "chart" at home that I was referring to.. I'll have to check on that one in the morning when I get back to the office...
Sabra
8:27
I will be honest on number one on pg 121 I am really not sure how to begin...
7:55
Is it because there is only ten ways to come up with how to have x1+x2 be less than 6? I am also on the same problem
Dayna
7:41
Thank you for the clarifications.  Follow-up question on #5..I am not understanding why it is the coefficient on z^10.
AvatarGreg Miller
6:39
Dayna.. regarding #5 on Page 46.. that one literally has to be handled on a case-by-case for x1 and x2... choose specific values for x1 and x2 that will satisfy the constraint and then solve the resulting problem over and over.. and add up all generating functions for each disjoint piece.  For instance (x1 =1, x2 =1.. then solve it)... (x1=1, x2=2).. then solve it.. etc.. etc... I made a chart of all my x1, x2 possibilities and then had a massive SUM of terms to get what they have in the back.. so no..magic bullet.. just case by case on the x1+x2<6 part. (notice its POSTIVE integers, so no zeros for x1, x2)
So that illustrates that the "complement rule".. or "look at the opposite question" approach can be profitable occasionally for generating functions as well.
6:35
Dayna.. original comment.. In #2 from Four Exercises.. we can consider the "opposite" counting problem which would be "... with no box empty"... then.. after we get the egf for that complementary problem.. subtract off the egf from the egf that we would have if there were NO CONDITIONS at all (which is e^(6z).. that gets us back to the egf for the stated problem "...with at least one box empty"...
6:31
The reason that the combination term is there is because the balls are distinguishable!  That's a total "gotcha" that Martin throws in there because usually balls are indistinguishable!!
Sabra
6:30
ahhh....thank you i was like where are 4 extra balls coming from :). So then that is to represent the ones not shown and left out
AvatarGreg Miller
6:29
Sabra.. the (34 choose 12) part is a typo.. should read (34 choose 8) in back of book
Sabra
6:27
On number 2 from the Five review problems, i worked it down to what i thought was the answer and then went to check...i had pieces right but had left off the (36,12) part. I am wondering why it says that when there would be 8 balls left over (this was with using 26 as r)....on using 34 as r  it had z^8/8! in the equation....is this to represent the 8 balls left behind?
6:25
I was iffy on #2 as well! Using that (e^z -1)^6 is only one item in the box so if we consider the boxes to be unlimited (e^z) so e^6z is all the ways with every possibility then taking away that possibility for there being one item would create that empty space....or at least that was my thought process...i  may be completely wrong on that...
AvatarGreg Miller
5:11
Chat opened at 5:11 PM...
Dayna
I am struggling with pg 64, Four Exercises, #2.  Are we first to address the "at least one empty" using the efg (e^z-1)^6 ?
5:11
I could also use a hint on page 64, question 5.  I believe the g.f for x3 would be 1/(1-z^3) and for x4, x5 would be (1/1-z).  For x1 and x2 I started with the possibility of x1 being 1,2,3,4 which would make x2 have to be between 0 and 5,  0 and 4, 0 and 3, and 1, respectively.  Then, I was thinking x1 and x2 could swap roles.  Am I on the right track?
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