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Combinatorics June 5 Discussion
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AvatarGreg Miller
10:27
Chat closed and archived at 10:27 PM.
8:16
Regarding "glue"... look back at the few times this has been done.. and you'll see that it's typically done to bind things that were distinguishable together... like Peter and Paul in Pg 15 #9 or my video round table count problem were I bound couples together... glue can be a great metaphor.. but it can fool us into creating scenarios where we unintentionally force order over the top of problems or force distinguishability over the top of problems when it isn't warranted.. when this happens.. unfortunately it is VERY difficult to uncouple and regroup to describe the overcount
8:10
Bethany, Comment at 5:59 regarding Pg 17 #5...   "Yes".
8:06
Bethany, Comment at 5:55 about Problem 10 on Page 16:  Maybe it help if thought of the answer as:  (2^1 ) + (2^2) + (2^3) + (2^4 - 2) + (2^5 - 2*(5C1) - 2) + (2^6 - 2*(6C2) - 2*(6C1) - 2).... which believe it or not.. is what I think is the more "natural" way to express the answer!
7:56
Describing "what specificially" you overcounted would be very, very difficult.... I'm not sure I could tell you that.  In other words, I'm not sure know WHAT to subtract off your overcount to correct it.. that is probably very complicated.. or.. maybe said more humbly.. I, for one, don't think I can deduce it.. but I do feel confident that the reason why the glue doesn't work is b/c it places distinguishability in an arena it wasn't meant to occupy.
7:54
Bethany.. regarding your comment at 5:15 about Pg 16, #5:  This is difficult to explain, but I'll try.  The 4 I's are indistinguishable.  You know this, of course.  But when you glue two of them together, then that glued block creates something distinguishable from something that was not meant to be.  You CAN distinguish two I's from the remaining  two individual I's that you didn't choose to glue.  Thus, you forced distinguishability where it wasn't deserved.  Since there are more ways to arrange a combination of distinguishable and indistingiushable objects than if they were still ALL indistinguishable.. you've imposed an overcount.
Dayna
7:08
Maybe the overlap is to account for possibly repeating a wedge, placing more I's together than just the two that are "glued" together.
Bethany
6:57
After reading the reasoning behind #1 on page 23 I am wondering if I need to change how I am wording my answers. Ex: First select the pole/space for each flag 18 choose 24 with repetition but unordered. Second order the flags on the pole/spaces. 24! Is how I would choose to explain this problem but the book is way different. Is anyone else struggling with how they are wording their answers?
6:15
Using the wedge method and 8 choose 2 with repetition after placing the glued "II" pair one of the possible 8 ways should account for all the possible combinations of I's. But it seems to actually have some extra ways that are subtracted out with 8 choose 4. I was just wondering what that overlap could be...
Dayna
6:06
On pg 16, #5 is it because you could either have all 4 I's together, or 1 I alone and the other 3 together, or two pair of I's or 1 pair of I's with the other two separate, making 4 possible combinations of I's?
Bethany
5:59
On Pg 17 # 5 we have to add an extra box to make the problem equal to 32 instead of less than. Is adding a ball to the extra box part of the process to equal 32 from being less than?
5:55
Page 16 number 10 I don't understand how to get the second half of the answer. I can see two distinguishable objects choose 1,2,3 but then I seem to get stuck when you have to select from both types of objects.There is some kind of change over there that I'm missing.
5:46
You're welcome Dayna.
Dayna
5:44
Thank you Bethany.  I was not making that connection but it makes perfect sense now!
Bethany
5:42
I thought that was because there was only one order possible so you can't use permutations.
Dayna
5:39
On page 15, #2 I am struggling with the fact that the letters must be in alphabetical order, but the back of the book uses the "unordered selection with repetition" formula.  Why do I feel like alphabetical order implies order?
AvatarGreg Miller
5:15
Chat open... 5:15 PM
Bethany
5:15
On page 16 #5 I understand how the back of the book came to the answer that it did, but I feel like you could also solve this by using "glue" and making an II group that can be placed on 8 wedges then placing the other two I's 8 choose 2 with repetition ways. This answer is slightly higher than the result from the book and I'm wondering what ways it is allowing that are not allowed or should be subtracted out by the 8 choose 4?
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