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Combinatorics July 1 Chat
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AvatarGreg Miller
11:22
Chat closed at 11:22 PM...
10:53
Yes
Sabra
10:07
so an is just all the dummy variables added together...and we manipulate that equation to get what to be able to plug in for them. Yep it just had not hit me yet
AvatarGreg Miller
10:02
Because A(n-1) = b(n-1)+c(n-1)+d(n-1).. the sequence has to end in "something"!
Cadey
9:39
The guest name is me btw... I forgot to put a name
Guest
9:38
In example 21. They go from b(n-1) to a(n-1) I understand it’s a dummy variable thing but I don’t follow it
Sabra
9:14
on example 21 how do we know that bn-1 + cn-1 is the same as an-1 -dn-1? that may be another duh question but it hasnt hit me yet
AvatarGreg Miller
8:33
The difference between 10 and 11 is that 10 is like Example 15 and #11 is not... #11 requires "immediately".. the difference is in the concept of "immediately after" vs. "anywhere after"...
8:30
..and  on that same theme.. I think possibly Example 20 or the dummy variable problem from videos could help with #8.. one thing that's a challenge is that sometimes you can think of a way to "force previous terms" into the logic that has nothing to do with "what happens at the start" or "what happens at the end".. like #10 from The Dozen... so when you try and create recurrence relations.. you really have to go through a LONG list of possible ways to model... VERY unlike generating function thinking where a lot of the problems are honestly a lot alike.
8:25
Personally, I think it may be best to think of #1 on Page 131 as being an extension of Example 21... which isn't how he attacks it in the "Back".
Bethany
7:48
Sabra I was also confused by the book on number 1. I got the initial conditions but when I tried to start from the beginning I could only think to start with two terms
Sabra
5:24
I am guessing that is the same idea applied to 11 and 12 from this section as well. I felt like 10 and 11 were asking me the same question honestly
5:22
On number 10 pg 132 is the 2^n-1 suppose to represent the fact we only have to other numbers we can choose from (0 or 2) to put in the slot next to 0?
5:21
I ran into trouble on number 8 on pg 132 i had 2an-1 + an-2
5:19
On number one on pg 131 i attempted to go from the beginning but i did not come up with what was in the back of the book....i don't think i began this problem the way I should have and was kind of confused by the explanation in the back
AvatarGreg Miller
4:41
Problem 12 on that page was also mentioned last night and it CAN be done by considering what we can do at the "start".. the boxes are distinguishable, so we can either put 2 or 3 or 4 boxes in the FIRST box... from there on out, we just have to put either r-2 or r-3 or r-4 balls in n-1 boxes.. and that's the definition of A(r-2, n-1) or A(r-3, n-1) or A(r-4, n-1) as the case may be.  To get the initial conditions, you just have to solve the problem separately case by cases for 0<=r<=4 and n=1 or 2 so that when you actually use the recurrence relation you'll be guaranteed to know what to put in for any term that would show up.. getting the initial conditions is probably the hardest part.  But again, this one is also weird because it is a bivariate recurrence relation and this is literally the only place in the homework where he pops one in.. so I'd consider this problem more a novelty.
4:20
Circling back around to #11 on Page 122 that was asked about last night.  He gives a really shrewd solution that I am not sure is very intuitive.  If you divide the list of distinct real numbers up in the problem, then you now have two piles of 2^(n-1) numbers.  So, the number of comparisons to get max/min of EACH pile is now A(n-1) by the definition of the sequence.  When you are done making the comparisons in those two piles you'll have a max from each pile and a min from each pile.  To get the OVERALL max, you have to make a comparison of the max's from each pile (that adds one to the count).. same for the OVERALL min (that adds one more to count).. so all together than makes A(n) = A(n-1) + A(n-1) + 1 + 1 = 2A(n-1) + 2.  That one's rough.. no way we get that unless we use the "half" trick.. it's not a "start", "end" type problem like some of these are...
3:54
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