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NZOlympiana

Tuesday, February 27, 2007. Here's an alternative (better) argument that the number of prepared sequences of length n is equal to n! Based on some ideas in Ilya's and Emily's papers. Let p n denote the number of prepared sequences of length n. Each such sequence arises from a prepared sequence of length (n-1) in one of two, mutually exclusive ways:. By inserting a copy of the largest element to the left of the leftmost occurrence in the sequence of length (n-1). Since p 1 = 1, p n = n! A convex quadrilat...

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NZOlympiana | nzimo.blogspot.com Reviews

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Tuesday, February 27, 2007. Here's an alternative (better) argument that the number of prepared sequences of length n is equal to n! Based on some ideas in Ilya's and Emily's papers. Let p n denote the number of prepared sequences of length n. Each such sequence arises from a prepared sequence of length (n-1) in one of two, mutually exclusive ways:. By inserting a copy of the largest element to the left of the leftmost occurrence in the sequence of length (n-1). Since p 1 = 1, p n = n! A convex quadrilat...

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1 nzolympiana

2 prepared sequences

3 by induction

4 0 comments

5 extra geometry iii

6 extra geometry ii

7 extra geometry i

8 geometry

9 have fun

10 hint combinatorics q8

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NZOlympiana | nzimo.blogspot.com Reviews

https://nzimo.blogspot.com

Tuesday, February 27, 2007. Here's an alternative (better) argument that the number of prepared sequences of length n is equal to n! Based on some ideas in Ilya's and Emily's papers. Let p n denote the number of prepared sequences of length n. Each such sequence arises from a prepared sequence of length (n-1) in one of two, mutually exclusive ways:. By inserting a copy of the largest element to the left of the leftmost occurrence in the sequence of length (n-1). Since p 1 = 1, p n = n! A convex quadrilat...

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nzimo.blogspot.com

1

NZOlympiana: Geometry

http://nzimo.blogspot.com/2006/02/geometry.html

Thursday, February 09, 2006. Our next assignment (coming soon I promise) will be on geometry. So you might want to spend some time doing a bit of reading of whatever materials you have available. At the EMAT 4600/6600. Posted by Michael Albert at 11:04 AM. Renegade mathematician, occasional poker player, and taking up bridge again. View my complete profile.

2

NZOlympiana: The Friendship Theorem

http://nzimo.blogspot.com/2006/01/friendship-theorem.html

Tuesday, January 24, 2006. Posted by Michael Albert at 9:46 AM. Renegade mathematician, occasional poker player, and taking up bridge again. View my complete profile.

3

NZOlympiana: Hint: Combinatorics Q5

http://nzimo.blogspot.com/2006/01/hint-combinatorics-q5.html

Friday, January 27, 2006. An obvious invariant: the number of rooms in which all three lights are on, or all three are off. We want to show that if this invariant is not zero it can be reduced. If it's not zero, we go to a "same-state" room and flick a switch. This might reduce the invariant. Hurrah we're done. If it doesn't then what happened? Can we actually have made the situation worse? What's the logical next room to try? What must eventually happen when we keep doing this? View my complete profile.

4

NZOlympiana: Hint: Combinatorics Q3

http://nzimo.blogspot.com/2006/01/hint-combinatorics-q3.html

Friday, January 27, 2006. First part: just be systematic and look for a solution with lots of structure. What we're trying to accomplish would seem to be helped by having relatively few entries in each row of any given label. So . So the maximum allowable number of edges is? On the other hand any pair of equal labels in any row produces an edge. So the minimum number of labels occurring is? Posted by Michael Albert at 7:46 AM. Renegade mathematician, occasional poker player, and taking up bridge again.

5

NZOlympiana: Hint: Combinatorics Q4

http://nzimo.blogspot.com/2006/01/hint-combinatorics-q4.html

Friday, January 27, 2006. This is a classic case where looking at much smaller examples will be a big help. So, go away and write down all. The ways of colouring:. The number 1 black or white. The numbers 1 and 2 black or white. The numbers 1, 2 and 3 black or white. And compute the associated numbers. When there's a unique odd value what is it? In general, how do the values change as we move through the list from left to right? Aim towards producing a simple. Problem was to go out and buy more keys!

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NZOlympiana

Tuesday, February 27, 2007. Here's an alternative (better) argument that the number of prepared sequences of length n is equal to n! Based on some ideas in Ilya's and Emily's papers. Let p n denote the number of prepared sequences of length n. Each such sequence arises from a prepared sequence of length (n-1) in one of two, mutually exclusive ways:. By inserting a copy of the largest element to the left of the leftmost occurrence in the sequence of length (n-1). Since p 1 = 1, p n = n! A convex quadrilat...

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