> ## Documentation Index
> Fetch the complete documentation index at: https://v1-learn.neoartd.my.id/llms.txt
> Use this file to discover all available pages before exploring further.

# Permutations and Combinations

## Permutations

Consider the possible arrangements of the letters **a**, **b**, and **c**.

> Bandingkan susunan yang memungkinkan dari huruf **a**, **b**, John **c**.

The possible arrangements are: **abc, acb, bac, bca, cab, cba**.

> Susunan yang memungkinkan adalah: **abc, acb, bac, bca, cab, cba**.

If the **order of the arrangement is important** then we say that each arrangement is a permutation of the three letters. Thus there are six permutations of the three letters.

> Jika **urutan susunan penting** maka kita katakan bahwa setiap susunan adalah permutasi dari tiga huruf. Dengan demikian ada enam permutasi dari tiga huruf.

An arrangement of n distinct objects in a specific order is called a permutation of the objects.

> Susunan n objek yang berbeda dalam urutan tertentu disebut permutasi dari objek.

<Tip>
  To determine the number of possibilities mathematically, one can use the multiplication rule to get: $3 \times 2 \times 1 = 6$ permutations.
</Tip>

## Permutation Rule

<Card>
  The arrangement of **n** objects in a specific order using **r** objects at a time is
  called a permutation of **n** objects taken **r** objects at a time. It is written as
  $_nP_r$ and the formula is given by

  $~~~~~_nP_r = \dfrac{n!}{(n-r)!}$.
</Card>

### Permutations - Example

<Card>
  How many different ways can a
  chairperson and an assistant
  chairperson be selected for a research
  project if there are seven scientists
  available?

  <br />

  Solution:

  Number of ways =
  $‎_7P_2 = \dfrac{7!}{(7-2)!} = \dfrac{7!}{5!} = 7 \times 6 = 42$.
</Card>

<Card>
  How many different ways can four
  books be arranged on a shelf if
  they can be selected from nine
  books?

  <br />

  Solution:

  Number of ways =
  $‎_9P_4 = \dfrac{9!}{(9-4)!} = \dfrac{9!}{5!} = 9 \times 8 \times 7 \times 6 = 3024$.
</Card>

## Combinations

Consider the possible arrangements of the letters **a**, **b**, and **c**.

> Bandingkan susunan yang memungkinkan dari huruf **a**, **b**, dan **c**.

The possible arrangements are: **abc, acb, bac, bca, cab, cba**.

> Susunan yang memungkinkan adalah: **abc, acb, bac, bca, cab, cba**.

If the **order of the arrangement is not important** then we say that each arrangement is the same. We say there is one combination of the three letters.

> Jika urutan susunan tidak penting, maka kita katakan bahwa setiap susunan adalah sama. Kita katakan ada satu kombinasi dari tiga huruf.

## Combinations Rule

<Card>
  The number of
  combinations of of **r** objects from
  **n** objects is denoted by $_nC_r$ and the formula is given by

  $~~~~~_nC_r = \dfrac{n!}{(n-r)!r!}$.
</Card>

### Combinations - Example

<Card>
  How many combinations of four
  objects are there taken two at a
  time?

  <br />

  Solution:

  Number of combinations =
  $‎_4C_2 = \dfrac{4!}{(4-2)!2!} = \dfrac{4!}{2!2!} = \dfrac{4 \times 3}{2 \times 1} = 6$.
</Card>

<Card>
  In order to survey the opinions of
  customers at local malls, a researcher
  decides to select 5 malls from a total of 12
  malls in a specific geographic area. How
  many different ways can the selection be
  made?

  <br />

  Solution:

  Number of combinations =
  $‎\_{12}C_5 = \dfrac{12!}{(12-5)!5!} = \dfrac{12!}{7!5!} = \dfrac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = 792$.
</Card>

<Card>
  In a club there are 7 women and 5 men. A
  committee of 3 women and 2 men is to be
  chosen. How many different possibilities
  are there?

  <br />

  Solution:

  Number of possibilities:
  (number of ways of selecting 3 women
  from 7) $\times$ (number of ways of selecting 2
  men from 5) =
  $‎_7C_3 \times ‎_5C_2 = \dfrac{7!}{(7-3)!3!} \times \dfrac{5!}{(5-2)!2!} = \dfrac{7!}{4!3!} \times \dfrac{5!}{3!2!} = 35 \times 10 = 350$.
</Card>

<Card>
  A committee of 5 people must be
  selected from 5 men and 8 women.
  How many ways can the selection
  be made if there are at least 3
  women on the committee?

  <br />

  Solution:

  The committee can consist of
  3 women and 2 men, or 4 women and 1
  man, or 5 women. To find the different
  possibilities, find each separately and
  then add them:
  $‎_8C_3 \times ‎_5C_2 + ‎_8C_4 \times ‎_5C_1 + ‎_8C_5 \times ‎_5C_0 = 56 \times 10 + 70 \times 5 + 56 \times 1 = 560 + 350 + 56 = 966$.
</Card>