# Combinatorics problem

A group of 3 couples has decided to start a dinner club. The first couple’s dinner table is rectangular with room for two people on either of the longer sides and room for one on either of the shorter sides. The second couple’s table is triangular, with room for two people on each side. The third couple’s table is circular. Up to rotations, how many different seating arrangements exist for each table?

1 2 3 4 ... n-1 n
1 2 3 4 ... 2 1
( n )
(m1, m2, ... mk)
this equals n! / (m1! * m2! * ... * mk!)
m1 + m2 + ... + mk = n

Let one seat be stationary at each different table. So then you have 5! which is the answer

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I am hoping someone here can be of some help.

For each table, think about how many arrangements are "equivalent" to a given one via some rotation. For example, at the triangular table, each possible seating arrangement is equivalent to exactly two others (there are two nontrivial ways to rotate an equilateral triangle into itself). Another way of saying this is that the size of an "equivalence class" of seating arrangements is 3. Thus, at the triangle, there are 6!/3 = 240 distinct ways of seating the couples.

For each table, think about how many arrangements are "equivalent" to a given one via some rotation. For example, at the triangular table, each possible seating arrangement is equivalent to exactly two others (there are two nontrivial ways to rotate an equilateral triangle into itself). Another way of saying this is that the size of an "equivalence class" of seating arrangements is 3. Thus, at the triangle, there are 6!/3 = 240 distinct ways of seating the couples.
So for the rectangle its 6! / 4 ? and the circle is 5! ?

Not quite; your answer for the circle is correct, but the total for the rectangular table is 6!/2. This is because the four sides of the table are not identical; two are distinguishable from the other two, so there's only one nontrivial way to rotate the table into itself.

Oh okay. Duh me :P lol. I should've caught that. Thanks for the help VKint.