Veritas prep counting problem

[kakz]

In how many different ways (relative to each other) can 8 friends sit around a square table with 2 seats on each side of the table?

[shankar.ashwin]

Don't think its any different from a regular circular arrangement. 7! ways IMO.

[GMATGuruNY]

In how many different ways (relative to each other) can 8 friends sit around a square table with 2 seats on each side of the table?

Let the 8 people be ABCDEFGH.

The number of ways to arrange N elements IN A LINE = N!.

When put in a line, the following qualify as distinct arrangements:

ABCDEFGH
BCDEFGHA
CDEFGHAB
DEFGHABC
EFGHABCD
FGHABCDE
GHABCDEF
HABCDEFG

But when put around a table, all of the above qualify as only ONE distinct arrangement, because the clockwise order in each is the same: A-B-C-D-E-F-G-H.
In all of the above:
B is directly to the right of A
C is directly to the right of B
D is directly to the right of C
E is directly to the right of D
F is directly to the right of E
G is directly to the right of F
H is directly to the right of G

Thus, the number of ways to arrange N people around a circular table is smaller than the number of ways to arrange the N people in a line:

The number of ways to arrange N elements around a CIRCULAR table = (N-1)!.

But with a SQUARE table, the number of distinct permutations increases.

Around a SQUARE table with TWO seats per side, each distinct clockwise ordering yields TWO distinct ways to place the elements:



The placements above are considered distinct because -- even though the clockwise ordering is the same -- different groupings are placed on each side of the table.
The result is that different people face each other.
For each distinct clockwise arrangement around a SQUARE table:
2 seats per side yields 2 possible groupings per side.
3 seats per side yields 3 possible groupings per side.
4 seats per side yields 4 possible groupings per side.
Each additional seat per side yields another possible grouping for each side of the table.
Thus, we get the following formula:

The number of ways to arrange N elements around a SQUARE table with M seats per side = M*(N-1)!

Thus, in the problem above, the number of ways to arrange the 8 people around a square table with 2 seats per side = 2*(8-1)! = 2*7! = 10,080.