A dinner party consisting of 5 couples, sit around a rectangular table

Question

A dinner party consisting of 5 couples, sit around a rectangular table, with ladies and gentlemen altering. The host and hostess each occupy one end of the table and their guests are arranged four on each side. What is the number of ways the party can be seated? (2 seating arrangements are considered different only when the positions of the people are different relative to each other.)

A. 72
B. 144
C. 288
D. 576
E. 1152

A. 72
B. 144
C. 288
D. 576
E. 1152

CrackverbalGMAT · Answer

Let us seat the host and Hostess at 1 end of the table. They can rearrange themselves in 2 ways.

There are 4 seats in which the men can seat alternatively. This can be filled in 4! = 24 ways.

The women will sit in the remaining 4 ways in 4! = 24 ways

Total Arrangements = 2 * 24 * 24 = 1152

Option E

Arun Kumar

ronaldoSuiiii · Answer

Actually it is mentioned in the question that the position of X is considered to be changed only if the person's position beside X is changed.
In this way, whereever host or hostess sits, if the person beside him/her doesn't change, the position would be considered to be unchanged.

Thus it should be only 24*24 considering the sides of host and hostess fixed ( i.e 4! seats of males and remaining 4! seats on females, 4!*4!).

= 24*24 
= 576
Answer D

varshas044 · Answer

Let’s see

We have total 10 people (5 couples) --- 5 males and 5 females

1) Head of the table will be 1 male and 1 female; they can only be arranged in one way since their position relative to each other will not change = 1 way
2) On either side of the rectangle (along the length) lets say that we have 2 males and 2 females on each side 
We have 4 males and 4 females left 
One side: Choices for arranging males = 4 x 3 ------ choices for arranging females = 4 x 3 
Other side: choices for arranging males = 2 x1 ---- choices for arranging females = 2x1

total ways = 1x4x3x4x3x2x1x2x1 = 576

Answer – D

AT1982 · Answer

F(4) M(4) F(3) M(3) 
hM hF
 F(2) M(2) F(1) M(1)

The alternating posn for M1 & F1 are constant.

Therefore 4! x 4! = 576

A dinner party consisting of 5 couples, sit around a rectangular table

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GMAT Problem Solving (PS) Questions

A dinner party consisting of 5 couples, sit around a rectangular table

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