This article “2 Wonderful Concepts – Arrangements and Selections” is about two important topics of Mathematics – **Permutation** and **Combination**. Before, moving on to these topics, let’s first take a look at one more associated term – **Factorial**.

## Factorial of a Number

The factorial of a whole number n, written as n!,’ is calculated by multiplying n by all the whole numbers less than it. For example, the factorial of 4 (written as 4!) is 24, because 4 × 3 × 2 × 1 = 24. Hence one can write 4! = 24.

Factorial is used to find out how many possible ways there are to arrange n objects.

For example, if there are 3 letters (A, B, and C), they can be arranged as ABC, ACB, BAC, BCA, CAB, and CBA.

It can be explained as:

Places to fit the 3 letters can be represented as ___ ___ ___

First letter A can be placed at any of the 3 places: A ___ ___ or ___ A ___ or ___ ___ A

Second letter B can be placed at any of the 2 remaining places: A B ___ or A ___ B or B A ____ or ____ A B or B ___ A or ___ B A

The third letter C will be left only with 1 vacant place.

Therefore, the number of ways 3 objects (letters in this case) can be arranged = 3 × 2 × 1 = 6 = 3!

The factorial function grows very fast. There are 10! = 3,628,800 ways to arrange 10 items.

n! is not defined for negative numbers and 0! = 1.

## Arrangements and Selections

Two important concepts in solving many problems are Permutation (Arrangement) and Combination (Selection). First, let’s see what is the difference between these two terms.

Let’s consider three objects – A, B, and C.

How many ways can you arrange two objects from the above three given objects?

The possible arrangements can be AB, BA, AC, CA, BC, and CB. There are 6 different ways 2 objects can be arranged from the given 3 objects. (Note that AB and BA are different as in AB, A comes first and in BA, B comes first. The same is true for CA & AC and CB & BC).

A permutation is an arrangement of r objects from given n objects and is denoted by ^{n}P_{r} or P(n, r). The permutation of r objects from given n objects is evaluated using the formula P(n, r) = n!/(n – r)!

In the above case, n = 3 and r = 2, P(n, r) = P(3, 2) = 3!/(3 – 2)! = 3!/1! = (3 × 2 × 1)/1 = 6.

Now, let’s see in how many ways two objects can be selected from these three given objects?

The possible selections will be AB, AC, and BC. There are 3 different ways 2 objects can be selected from the given 3 objects. (Note that BA is the same as AB, as BA and AB both means selecting A and B. The same is true for CA & AC and CB & BC).

A combination is a selection of r objects from given n objects and is denoted by ^{n}C_{r} or C(n, r). The combination of r objects from given n objects is evaluated using the formula C(n, r) = n!/[r!(n – r)!]

In the above case, n = 3 and r = 2, C(n, r) = C(3, 2) = [3!/2!(3 – 2)!] = [3!/(2!)1!] = (3 × 2 × 1)/((2 × 1) × 1) = 3.

## Difference Between Permutation and Combination

The following examples illustrate the difference between Permutation and Combination.

Permutation | Combination |

Arrangement of cuisines in a menu in a restaurant. | Selection of cuisines from a menu in a restaurant. |

Arrangement of flowers in a bouquet. | Selection of flowers from a bunch of flowers. |

Arrangement of team players in a line. | Selection of team players from a group of players. |

Let’s consider the following example. There are 20 players and the selectors want to select 11 players from these 20 ways. How many different teams of 11 players can be formed? Here n = 20 and r = 11. Since selectors want to select the players, therefore, we will use the combination.

The number of distinct teams that can be formed = C(20, 11) = 20!/(11! × (20 – 11)!) = 20!/(11! × 9!)

= (20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12 × 11!)/(11! × 9!)

= (20 × 19 × 18 × 17 × 16 × 15 × 14 × 13 × 12)/(9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1)

= 167,960

Now, we want to know in how many ways these 11 players can stand in a line. In this case, we will use permutation, since we want the players to be arranged in a line. Here n = 11 and r = 11 (Since all 11 players are to be arranged in a line). The different lines (arrangements) that can be formed is P(11, 11) = 11!/(11 – 11)! = 11!/0! = 11! (0! = 1) = 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 39,916,800.

Permutation and combination are widely used in solving tricky mathematics puzzles. Find some real-world problems on permutation and combination here: