Sam usually takes one main course and a drink. Today he has the choice of burger, pizza, hot dog, watermelon juice and orange juice. What are all the possible combinations that he can try?
There are 3 main course choices and 2 drink choices. We multiply to find the combinations. \(3\times 2=6\).
Thus Sam can try 6 combinations.
In this minlesson, let's learn about permutation, arrangement, combination, collections, factorials, worksheets, formulas, examples
Lesson Plan
What Are Permutations And Combinations?
The process of selecting r items out of n items is called permutations and combinations.
Differences between permutations and combinations
Permutations  Combinations 

Arranging r items out of n items.  Selecting r items out of n items. 
There are 5 students in a group. The teacher is going to pick 3 students for a prize. The first person she picks will get the 1st prize, the second student, the 2nd prize and 3rd prize, the third. 
There are 5 students in a group. The teacher is going to pick 3 students for a prize. All get the same prize. 
Solution: 60 ways  Solution: 10 ways 
Do you know how we got the solution?
To get on the hook of it, let's learn to find the factorials.
Factorials
The factorial function is a product of the series of descending natural numbers.
\(\begin{align} n! &= 1.2.3.4.…n\\ &= \text{Product of the first n positive integers}\\&= n(n1)(n2)…..(3)(2)(1)\end{align}\)
cuemath.com/geometry/quadrant/ 
Examples:
\(3! = 3\times 2\times 1\)
\(4! = 4\times3\times 2\times 1\)
\(4! = 4\times 3!\)
\(n! = n \times (n  1)!\) 
We use the factorials in finding the permuations and the combinations.
Permutations And Arrangements
A permutation is an ordered arrangement of outcomes. It is an ordered combination.
It is represented in the following ways:
Picking 3 students in order from a group of 5 to give away prizes based on their performance in athletics.
To give the first prize you pick in 5 ways.
To give the second, you pick in 4 ways.
To give the third, you pick in 3 ways.
Thus for getting the number of ways for arranging 3 persons from 5 persons, we multiply the options available to us.
We do it in 5 . 4 . 3 ways.
By analyzing this way, we find that this could be done in \(5! \times 4 ! \times 3! = 60 \) ways.
This is generalized using the formula:
^{n}P_{r} = \(\dfrac{{n!}}{{(n  r)!}}\) 
Examples are used to show permutation with repetition and permutation without repetition.
Example 1: There are 3 distinct letters in the word CAT.
The arrangements of the letters of that word is
\[3P_3 = 3! = 6 \text{ways}\]
CAT ATC TCA
CTA ACT TAC
The letters in CAT are not repeated.
Example 2: Consider the 4letter word CUCUMBER.
U and C repeated twice.
The arrangements of the letters of that word is
\[8P_8 = 8! = 40320 \text{ways}\]
The arrangements of the letters C is
\[2P_2 = 2! = 2 \ \text{ways}\]
The arrangements of the letters U is
\[2P_2 = 2! = 2 \ \text{ways}\]
\(\therefore\) We divide \(8P_8\) and \(2P_2\times 2P_2\) and get the final arrangement. i.e
\[\dfrac{40320}{4}=10080 \text{ways}\]
Try to play with the objects in the following simulation and understand how permutations happen.
Combinations And Collections
A combination is an arrangement of outcomes in which the order does not matter.
It is r objects chosen from n objects and the order is not important.
It is represented in the following ways:
Example: Select 3 books from 5 books on the shelf.
We can choose it in \(5P_3 = 60 \) ways.
The possible ways the books could be selected doesn't require an order as we can choose any at random.
Thus we divide 60 by 3 !. i.e. \(\dfrac{60}{6} = 10\) ways.
^{n}C_{r} = \(\dfrac{{n!}}{{r!(n  r)!}}\) 
Try to play with the objects in the following simulation and understand how combinations happen.
.
 When the order doesn't matter, it is a Combination.
 Hint : group, sample, selection
 When the order does matter it is a Permutation.
 Hint :arrangement, schedule, order
Applications of Permutations And Combinations
Permutations : Allocation of phone numbers according to the country code, area codes, Assigning Car number plates according to the state codes, setting up of passwords and passcodes to maintain the security credentials.
Combinations: selecting nominees for various positions, selecting the lottery numbers.

The formulas for finding permutations and combinations are :

nPr = \(\dfrac{{n!}}{{(n  r)!}}\)

nCr = \(\dfrac{{n!}}{{r!(n  r)!}}\)

nCr = \(\dfrac{nPr}{r!}\)
Important Topics on Permutations and Combinations
Given below is the list of topics that are closely connected to permutations and combinations. These topics will also give you a glimpse of how such concepts are covered in Cuemath.
 Practical Counting Situations
 Examples  Practical Counting Situations
 Permutations
 Examples  Permutations as Arrangements
 Combinations
 Examples  Combinations as Selections
Solved Examples
Example 1 
Kathleen, is teaching 5 lessons to her class. In how many different orders could the lessons be taught?
Solution
Since the orders matter, we need to find the permutation of the lessons.
So choosing 1 lesson is done in \(5P_15\) ways.
\[\begin{align} 5P_5&= \dfrac{5 ! }{1 !}\\&=5\times 4\times 3\times 2\times 1\\&=120\end{align}\]
\(\therefore\) 120 orders 
Example 2 
Patricia has to choose 5 marbles from 12 marbles. In how many ways can she choose them?
Solution
Patricia has to choose 5 out of 12 marbles.
Order doesn't matter here.
Thus she can choose it in \(12C_5\) ways.
\[\begin{align}12C_5&= \dfrac{12!}{5 ! \times(125) !}\\ &= \dfrac{12!}{5 ! \times 7!}\\&= \dfrac{12\times 11\times 10\times 9\times 8\times 7 !}{5 ! \times 7 !}\\&= \dfrac{12\times 11\times 10\times 9\times 8}{5 ! }\\&= 792\end{align}\]
\(\therefore\) 792 ways 
Example 3 
How many diagonals can be formed in the polygon?
Solution
To form a diagonal, we need 2 nonadjacent vertices.
(\(\because\) 2 adjacent vertices will form a side of the polygon and not a diagonal).
The total number of ways of selecting 2 vertices out if n is \(nC_2\).
This number also contains the selections where the 2 vertices are adjacent.
Thus, the total number of diagonals is \(nC_2  n\)
\(\begin{align}nC_2  n\\ &= \dfrac{n(n1)}{2}  n\\ &= \dfrac{n(n3)}{2}\end{align}\)
\(\therefore\)\(\dfrac{n(n3)}{2}\) 
 How many squares are there on a standard \(8 \times 8\) chessboard?
 At a party, everyone shook hands with everybody else. There were 66 handshakes. How many people were at the party?
Interactive Questions
Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.
Let's Summarize
The minilesson targeted the fascinating concept of permutations and combinations. The math journey around permutations and combinations starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that not only it is relatable and easy to grasp, but also will stay with them forever. Here lies the magic with Cuemath.
About Cuemath
At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!
Through an interactive and engaging learningteachinglearning approach, the teachers explore all angles of a topic.
Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.
Frequently Asked Questions (FAQs)
1. What are factorials?
Factorials are the product of the first n positive integers.
\(\begin{align}
n!&= 1.2.3.4.…n\\
&= \text{Product of the first n positive integers}\\
&= n(n1)(n2)…..(3)(2)(1)\end{align}\)
\(5! = 5\times 4\times 3\times 2\times 1\)
2. What is 0! ?
\(0 ! = 1\)
3. What is the difference between permutations and combinations?
Permutations are arrangements made in order, whereas combinations are the choices made never minding the order.
Permutations are used to list, however combinations are used to group.
4. When to use permutations and combinations?
When we need to choose r objects from n objects, we use permutations and combinations.
Permutations are ordered combinations.