**No seating restrictions.****A****and**B**sit together?****4****men and****4****women and no****2****men or****2****women can sit together?****5****men must sit together?****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.**

## Expert Answer

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** = $n!$

\[=8!\]

\[=8\times 7\times 6\times 5\times 4\times 3\times 2\times 1\]

\[= 40,320\space Possible\space Ways \]