Perms and Combs (1 Viewer)

[Aerath]

Show that 2n persons can seat themselves at 2 identical round tables, n persons at each, in (2n)!/n^2 ways.

Thanks in advanced.

 

[lolokay]

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?

 

[gurmies]

EDIT: BEATEN BY LOLOKAY -.-