Four married couples have bought $8$ seats in the same row for a concert.

In how many different ways can they be seated

$\left(a\right)$ with no restrictions?

$\left(b\right)$ if each couple is to seat together?

$\left(c\right)$ if all the men sit together to the right of all the women?

Step 1:Finding the number of ways four married couple seated with no restriction:

Given,

Total number of people $8$.

Total number of seats $8$

If there is no restriction then

First seat have $8$ possible way.$\Rightarrow n$ [since eight persons are there]

Second seat have $7$possible way. $\Rightarrow n-1$[since there is only seven person already one person is have seat]

Third seat have $6$ possible ways$\Rightarrow n-2$[since there is only six person already two person is have seat]

likewise

seventh seat $2$ ways $\Rightarrow n-6$[since there is only two person already six person is have seat]

Eighth seat have $1$ way.$\Rightarrow n-7$[since there is only one person already seven person is have seat]

So,

The number of ways be $8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$ ways.

Alter:

We know that,

If there is no restriction in seating arrangement the number of the possible ways $=n!$

Here $n=8$

So,

The number of ways be $8!=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1=40320$ ways.

Hence, the number of ways four married couples have bought $8$ seats in the same row for a concert with no restriction is $40320$ ways.

Step 2: Finding the number of ways four married couples have bought $8$ seats in the same row for a concert if each couple is to seat together:

If there are four married couples.

As a couple they are calculated as one . Couples have no restriction

We know that,

If there is no restriction in seating arrangement the number of the possible ways $=n!$

So,

The four married couples can be arranged in $4!=4\times 3\times 2\times 1=24$ways

Each pair has $2!$ ways.[first seat placed for men and second seat for women or first seat placed for women and second seat for men]

So $4$ pairs has $2!\times 2!\times 2!\times 2!$

So the total number of ways $=4!\times 2!\times 2!\times 2!\times 2!$

$$\begin{gathered} =4!\times 2!\times 2!\times 2!\times 2! \\ =24\times 2\times 1\times 2\times 1\times 2\times 1\times 2\times 1 \\ =384 \end{gathered}$$

Hence, the number of ways four married couples have bought $8$ seats in the same row for a concert if each couple is to seat together is $384$ ways.

Step 3: Finding the number of ways four married couples have bought $8$ seats in the same row for a concert if all the men sit together to the right of all the women:

Four married couples means $4$ men and four $4$ women.

The number of arrangement of these $4$ men is $4!=4\times 3\times 2\times 1=24$ ways [If there is no restriction in seating arrangement the number of the possible ways $=n!$]

The number of arrangement of these $4$ women is $4!=4\times 3\times 2\times 1=24$ ways

if all the men sit together to the right of all the women is $1!$.

So the total number of ways $=4!\times 4!\times 1!$

$$\begin{gathered} =4\times 3\times 2\times 1\times 4\times 3\times 2\times 1\times 1 \\ =24\times 24\times 1 \\ =576 \end{gathered}$$

Hence, the number of ways four married couples have bought $8$ seats in the same row for a concert if all the men sit together to the right of all the women is $576$ ways.