| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Proposed by Norman J. Finizio and James T. Lewis, University of Rhode
Island, Kingston.
A circular arrangement of the integers { 1, 2, ..., 2n } is "balanced"
if a+b = a'+b' whenver a and b are adjacent and a' and b' are
diametrically opposite a and b. How many balanced circular
arrangements of { 1, 2, ..., 2n} are there? [As usual, rotations are
not counted as different.]
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 1712.1 | an answer | HERON::BUCHANAN | The was not found. | Fri Jan 29 1993 08:02 | 18 |
Assume that reflections *are* counted as different, since that is what is hinted. Re-express the constraint as: a-a' = -(b-b'), where a & b are adjacent. We can continue comparing adjacent pairs like this all the way round the circle. When we get half-way round, we see there is no solution if n is even. So n is odd. We see that each of the n pairs of opposed numbers differs by the same amount, K. Since there are n pairs, this is only possible if K = n. Now fix the position of 1, and arrange the other n-1 pairs how one one likes. So: n even => 0 solutions n odd => (n-1)! solutions Andrew. | |||||