 # In how many ways can 8 people be seated in a row if: 1. No seating restrictions.
2. A and sit together?
3. 4 men and 4 women and no 2 men or 2 women can sit together?
4. 5 men must sit together?
5. 4 married couples must sit together?

The aim of this problem is to introduce us to probability and distribution. The concepts required to solve this problem are related to introductory algebra and statistics. Probability is just how plausible something is to occur. Whenever we are uncertain about the result of an event, we can look into the probabilities of how likely the results are to occur.

Whereas a probability distribution is a mathematical equation that presents the probabilities of events of various probable outcomes for experimentation.

According to the problem statement, we are given a total number of $8$ people sitting in a row, so let’s say $n=8$.

Part a:

The number of ways, $8$ people can be seated without restrictions $=n!$.

Therefore,

Total number of ways $=n!$

$=8!$

$=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$

$=40,320\space Possible\space Ways$

Part b:

Since $A$ and $B$ must sit together, they become a single block, so $6$ other blocks plus $1$ block of $A$ and $B$ makes $7$ positions to catch up with. Thus,

$=7!$

$=7\times 6\times 5\times 4\times 3\times 2\times 1$

$=5,040\space Possible\space Ways$

Since $A$ and $B$ are separate, so $A$ and $B$ can be seated as $2! = 2$.

Thus, the total number of ways become,

$=2\times 5,040=10,080\space Ways$

Part c:

Assume any of the $8$ persons on the first position,

First positon $\implies\space 8\space Possible\space Ways$.

Second positon $\implies\space 4\space Possible\space Ways$.

Third positon $\implies\space 3\space Possible\space Ways$.

Forth positon $\implies\space 3\space Possible\space Ways$.

Fifth positon $\implies\space 2\space Possible\space Ways$.

Sixth positon $\implies\space 2\space Possible\space Ways$.

Seventh positon $\implies\space 1\space Possible\space Ways$.

Eighth positon $\implies\space 1\space Possible\space Ways$.

Now we are going to multiply these possibilities:

$=8\times 4\times 3\times 3\times 2\times 2\times 1\times 1$

$= 1,152 \space Possible\space Ways$

Part d:

Let’s assume that all the men be a single block plus $3$ women still individual entities,

$=4!$

$=4\times 3\times 2\times 1$

$=24\space Possible\space Ways$

Since there are $5$ individual men, so they can be seated as $5!=120$.

Thus, the total number of ways becomes,

$=24\times 120=2,880\space Ways$

Part e:

$4$ married couples can be arranged in $4!$ ways. Similarly, each couple can be arranged in $2!$ ways.

The number of ways = $2!\times 2!\times 2!\times 2!\times 4!$

$=2\times 2\times 2\times 2\times 4\times 3\times 2\times 1$

$=384\space Possible\space Ways$

## Numerical Result

Part a: $40,320\space Ways$

Part b: $10,080\space Ways$

Part c: $1,152\space Ways$

Part d: $2,880\space Ways$

Part e: $384\space Ways$

## Example

Let $4$ married couples be seated in a row. If there are no restrictions, find the number of ways they can be seated.

The number of possible ways in which $4$ married couples can be seated without any restriction is equal to $n!$.

Therefore,

The number of ways = $n!$

$=8!$

$=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$

$= 40,320\space Possible\space Ways$