The purpose of this question is to understand the concepts of **permutations** and **combinations** for evaluating a different number of possibilities of a given event.

The **key concepts** used in this question include **Factorial, ****Permutation and ****Combination. **A **factorial is a mathematical function** represented by the **symbol !** that operates only on the positive integers. In fact, if n is a positive integer, then its factorial is **the product of all positive integers smaller than or equal to n**.

Mathematically:

**\[n! = n \cdot (n-1) \cdot (n-2) \_.\_ .\_ 3 \cdot 2 \cdot 1 \]**

For example, $4! = 4.3.2.1$ and $10! = 10.9.8.7.6.5.4.3.2.1$

**Permutation is a mathematical function** used to numerically calculate different **number of arrangements** of a certain subset of items when **order of arrangements is unique and important.**

If $n$ is the number of total elements of a given set, $k$ is the number of elements used as a subset to be arranged in a certain order, and $!$ is the factorial function, then **permutation can be represented mathematically** as:

**\[P(n,k) = \frac{n!}{(n-k)!} \]**

There is **another function** used to find the number of such possible subset arrangements **without giving attention to the order of the arrangements** rather than focusing on the subset elements only. Such a function is called a **combination**.

A **Combination** is a mathematical function used to numerically calculate the number of **possible arrangements** of certain items in a case where the **order of such arrangements is not important**. It is most commonly applied in solving problems where one has to make teams or committees or groups out of total items.

If $n$ is the number of total elements of a given set, $k$ is the number of elements used as a subset to be arranged in a certain order, and $!$ is the factorial function, the **combination can be represented mathematically as:**

**\[C(n,k) = \frac{n!}{k!(n-k)!}\]**

**Permutations and combinations** are often confused with one another. The **main difference** is that **permutations are order sensitive while combinations are not**. Let’s say that we wish to create **a team of 11 players out of 20**. Here the order in which 11 players are selected is irrelevant, so it’s an example of a combination. However, if we were to seat those 11 players on a table or something in a certain order, then it would be an example of permutation.

## Expert Answer

This question is **order sensitive**, so we will **use permutation** formula:

\[P(n,k) = \frac{n!}{(n-k)!}\]

**Substituting $n = 5$ and $k = 5$** in above equation:

\[P(5,5) = \frac{5!}{(5-5)!}\]

\[P(5,5) = \frac{5.4.3.2.1}{(0)!}\]

\[P(5,5) = \frac{120}{1}\]

\[P(5,5) = 120\]

## Numerical Result

There are **120 different orders** in which five runners can finish a race if no ties are allowed.

## Example

In how many **different ways can the letters A, B, C and D be arranged** to form two letter words?

**Recall the formula of permutations:**

\[P(n,k) = \frac{n!}{(n-k)!}\]

**Substituting $n = 4$ and $k = 2$ **in the above equation:

\[P(4,2) = \frac{4!}{(4-2)!} = \frac{4!}{(4-2)!} = \frac{4.3.2.1}{(2.1)!}\]

\[P(5,5) = 12\]