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Perms and Combs (1 Viewer)
- Thread starter Aerath
- Start date
2nCn = (2n)!/(n!)^2 to choose the n people for each table
round tables, so for each table if you hold one person constant there are then n-1 people to be arranged, which can be done in (n-1)! at each table
overall: (2n)!/(n*(n-1)!)2 * (n-1)!2
= (2n)!/n2
edit: another way, is just to arrange the 2n people at the 2n seats, as if the 2 tables went along in a row - (2n)! ways. then, as the tables are actually round, and so there is no start at each table, we divide by n for each table
i feel however, that we should divide by 2, as the tables are identical?
round tables, so for each table if you hold one person constant there are then n-1 people to be arranged, which can be done in (n-1)! at each table
overall: (2n)!/(n*(n-1)!)2 * (n-1)!2
= (2n)!/n2
edit: another way, is just to arrange the 2n people at the 2n seats, as if the 2 tables went along in a row - (2n)! ways. then, as the tables are actually round, and so there is no start at each table, we divide by n for each table
i feel however, that we should divide by 2, as the tables are identical?
Last edited:
gurmies
Drover
EDIT: BEATEN BY LOLOKAY -.-