Question

$\text{There are}2\text{indian couples,}2\text{american couple and}1\text{unmarried person}$
$\begin{array}{cccc}& \text{List- I}& & \text{List-II}\\ & & & \\ \left(I\right)& \text{The total mumber of ways in which they can sit}& \left(P\right)& 5760\\ & \text{in a row such that an indian wife and an american}& & \\ & \text{wife are always on either side of the unmarried person}& & \\ & & \left(Q\right)& 24230\\ \left(II\right)& \text{The total mumber of ways in which they can sit}& & \\ & \text{in a row such that the unmarried man always}& & \\ & \text{occupy the midddle position}& & \\ & & \left(R\right)& 40320\\ \left(III\right)& \text{The total mumber of ways in which they can sit}& & \\ & \text{around a circular table such that indian wife and}& & \\ & \text{american wife are on either side of unmarried person}& & \\ & & \left(S\right)& 1920\\ \left(IV\right)& \text{The total number of ways in which they can sit in}& & \\ & \text{a row such that all couples sit together}& & \\ & & \left(T\right)& 28410\end{array}$

$2)$ Which of the following is only INCORRECT combination?

• A. $I\to \left(R\right)$
• B. $II\to \left(P\right)$
• C. $III\to \left(P\right)$
• D. $IV\to \left(S\right)$

Answer 1

The correct option is B $II\to \left(P\right)$

For case $I\to$:
An indian wife and an american wife can be selected in $={}^2{C}_1\times {}^2{C}_1=4$ ways
Now, we need to make arrangements in which both of these indian and american wives sit around the unmarried person. There are two possibility as (IUA),(AUI) (I, U and A represent indian wife, unmarried person and american wife respectively).
we will consider (IUA or AUI) as a one element and there are $6$ elements (persons in this case) remain. That can be arranged in $7!$ ways
So, total number of ways $=4\times 2\times 7!=8!$

For case $II$
Number of ways in which the unmarried person can be arranged $=1$
Now, other $8$ persons can be arranged in $8!$ ways.
So, total number of ways = $8!$

For case $III$
This case is same as the case $I$, but here the arrangement is circular.
We will consider (AUI or IUA) as a one object and there are $6$ objects (persons) remain. so, the number of ways of arranging the $7$ objects in a circluar table $6!$
So, total number of ways $=4\times 2\times 6!=5760$

For case $IV$
Here, we have to arrange $4$ couples and $1$ unmarried person. The total number of ways $=5!\times (2!{)}^4=1920$ (a couple can be internally arranged in $2!$ ways)