Chapter 8.5, Problem 25E

Solution

(a)

The total number different ways in which four couples can sit.

40320 ways

Given:

There is no sitting restriction.

Calculation:

Number of ways in which 4 couples can be arranged with no sitting restrictions will be ${}_8{P}_8$ .

  $\begin{array}{l}=\frac{8!}{(8-8)!}\\ =40320\end{array}$

(b)

The total number different ways in which four couples can sit.

384 ways

Given:

The two members of each couple will sit together.

Calculation:

Now considering two members of a couple as 1, there are 4 persons and 4 seats available.

Number of ways in which 4 couples can be arranged with no sitting restrictions will be ${}_4{P}_4$

Consider the couples in a specific arrangement when male sits to the right of the lady. Now other arrangements would be that either one of the couple sits with male to the left of the lady or two couple with male on the left and so on.

This leads to ${}_4{C}_0{\times }_4{C}_1{\times }_4{C}_2{\times }_4{C}_3{\times }_4{C}_4$ ways of arranging couples among themselves

Hence, from Fundamental Counting Principle,

  $\begin{array}{c}{}_4{P}_4\times \left({}_4{C}_0{\times }_4{C}_1{\times }_4{C}_2{\times }_4{C}_3{\times }_4{C}_4\right)=\frac{4!}{0!}\times \left({\displaystyle \sum _{i=0}^4{}_4{C}_i}\right)\\ =24\times \left({\displaystyle \sum _{i=0}^4\frac{4!}{\left(4-i\right)!i!}}\right)\\ =24\times \left(1+4+6+4+1\right)\\ =24\times \left(16\right)\\ =384\end{array}$