Perms and Combs (2 Viewers)

Aerath · May 3, 2009

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 · May 3, 2009

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 · May 3, 2009

$2nCn.(n-1)!.(n-1)!$

$Explanation: 2nCn \,\, ways \,\, to \,\, choose \,\, n \,\, people \,\, from \,\, 2n. \,\, Multiply \,\, by \,\, (n-1)! \,\, to \,\, to \,\, arrange \,\, these \,\, n \,\, people \,\, around \,\, the \,\, first \,\, table. \,\, Multiply \,\, by \,\, (n-1)! \,\, to \,\, arrange \,\, the \,\, rest \,\, around \,\, the \,\, second \,\, table.$

$2nCn.(n-1)!.(n-1)! = \frac{(2n)!}{n!.n!}.(n-1)!(n-1)!$

$= \frac{(2n!)}{n(n-1)!.n(n-1)!}.(n-1)!(n-1)!$

$= \frac{(2n)!}{n^{2}}$

$Q.E.D.$

EDIT: BEATEN BY LOLOKAY -.-

Perms and Combs (2 Viewers)

Aerath

Retired

lolokay

Active Member

gurmies

Drover

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