1145 LearnersLast updated on June 5th, 2025
Permutation and Combination
There are different concepts in math when it comes to counting or sorting the data in groups. The most common concepts are permutation and combination. In this topic, we will discuss more about permutations and combinations, their formulas, and their differences.
What are Permutation and Combination?
The methods that we use in counting how many outcomes are possible in a situation are permutation and combination. Factorials are an important concept used to understand the concepts of permutations and combinations. The product of the first n natural numbers is n!
Permutations
Permutations are arrangements of a definite order of a number of events. It is the number of ways a set of objects can be arranged. In permutation order and the arrangement of data is important. There are two types of permutation:
- Permutations with repetition
- Permutations without repetition.
Combinations
In combination the number of ways an event can be arranged and here the order doesn't matter. Here, the types of the same kind of things are sorted. In combination, the order and the arrangement of the data do not matter.
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Difference between permutations and combinations
Now let’s learn a few differences between permutations and combinations
| Permutation | Combination |
| In permutation, the order of events matters. | In combination, the order is not required |
| If the events are different we use permutation | Used when we want to select the items from a group without considering order. |
| Permutation is calculated using the formula, P(n, r) = n! / (n - r)! |
Combination is calculated using the formula, C(n, r) = n! / [r! (n - r)!] |
| For example, The sequence of a number of a lock Selecting individuals for a team by position |
For example, Numbers for winning the lottery Selecting children |
Derivation of Formulas for Permutation and Combination
The permutation formula is P(n, r) = n! / (n - r)!. The combination formula is C(n, r) = n! / [r! (n - r)!]. Let us now learn, how these formulas for permutations and combinations are obtained.
Derivation of Permutation Formula
Permutation is the arrangement of objects or events in an ordered manner. To obtain the formula for permutation, we will consider the given scenario.
When selecting r objects from n objects, the first object can be chosen in n ways. The second object can be chosen from the remaining (n -1) objects, for the third object we are left with (n - 2) objects.
So, P(n, r) = n × (n -1) × (n - 2) × ….. × (n - r + 1),let’s consider this as 1
As this is a part of factorial notation. The factorial of a number n! is:
n! = (n -r) × (n - r - 1) × (n - r - 2) × ….. 3 × 2 × 1, let’s consider this as 2
Multiplying and dividing 1 with 2
That is (n × (n -1) × (n - 2) × ….. × (n - r + 1)) ((n -r) × (n - r - 1) × (n - r - 2) × ….. 3 × 2 × 1) / (n -r) × (n - r - 1) × (n - r - 2) × ….. 3 × 2 × 1
So, P(n, r) = n! / (n - r)!
Derivation of Combination Formula
Combination is the way of selecting r objects without considering order. Now let's see what and how the formula of combination derived
As we know the formula for permutation is P(n, r) = n! / (n - r)!, it is for r objects out of n objects with order consideration.
As ordering is not considered in combination, let's remove the ordering. Then the number of ways to arrange the selected r object among themselves is r!,
Since, C(n, r) is the ratio of the total number of permutations to the number of ways to arrange r different objects
So, C(n, r) = P (n, r) / r!
That is C(n, r) = n! / (n - r)! r!
Real-world applications of Permutation and Combination
Permutations and combinations are the techniques we use in math to calculate the order of events, and it is used in different fields. Let’s learn a few real-life applications of permutations and combinations.
- Scheduling and planning events: To plan events in order we use permutation that can be performed or to schedule meetings and shifts.
- Permutations are used in cryptography and security to estimate the possible keys and to create a secure code
- In lottery and gaming, we use both permutations and combinations. For picking the set of winning numbers in the lottery we use combinations and
- In cooking, both permutations and combinations are used for mixing the ingredients
Common Mistakes and How to Avoid Them in Permutation and Combination
Now let’s learn a few common mistakes that students tend to make when working with permutation and combination.
Mistake 1
Confusing permutations and combinations
Students tend to confuse the concept of permutations and combinations as they get confused when the order matters. To avoid this error they should understand that the order matters in permutation, and it doesn’t matter in combinations.
Mistake 2
Incorrect application of formula
Sometimes students use incorrect values in the formula, misplacing factorial or confuse the formula. So they need to write the formula step-by-step and check whether it's correct or not in each step.
Mistake 3
Ignoring the restrictions in questions
Ignoring the conditions in the questions such as no two adjacent must include a particular element, etc. can lead to counting errors. So students should read the questions carefully and mention the restrictions. It is important to list all the conditions, then we have to adjust the formula accordingly.
Mistake 4
Incorrect counting cases for at least and at most
Confusing with conditions like at least and at most is common among kids, and it can lead to overcounting or undercounting. So, to avoid them, they understand it and count it accordingly. So to find it, first we need to find the total number of cases. Then, we have to find the count of the none case. Then find the difference between the total outcomes and the outcomes with none.
Mistake 5
Overcounting arrangement with repeated elements
Considering the repeated items as a separate entity and using n! Directly will lead to wrong counting. So adjust the count by dividing the factorial of each group of identical elements
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Solved Examples of Permutation and Combination
Problem 1
A student has 5 different books. In how many orders can the books be arranged on a shelf?
The books can be arranged in 120 ways
Explanation
As all 5 books are different, we use permutation to find the number of arrangements.
P(n, r) = n! / (n - r)!
Here, r = 5 and n = 5
So, P(5, 5) = 5! / (5 - 5)!.
= 5! / 0!
5! = 5 × 4 × 3 × 2 × 1 = 120
0! = 1
So, 5! / 0! = 120.
Problem 2
A club has 10 members. In how many ways can a committee of 4 members be chosen?
In 120 ways we can select a committee of 4 members.
Explanation
As the order is not considered when selecting the committee members we use combinations.
C(n, r) = n! / (n - r)! r!
Here, n = 10 and r = 4
So, C(10, 4) =10! / (10 - 4)! 4!
= 10! / 4! × 6!
10! = 10 × 9 × 8 × 7 × 6!
Canceling 6! From numerator and denominator
= 10 × 9 × 8 × 7 × 6! / 4! × 6!
= 10 × 9 × 8 × 7 / 4!
4! = 4 × 3 × 2 × 1 = 24
So, 5040 /24 = 210.
Problem 3
How many ways can the letters of the word “APPLE” be arranged, noting that the letter P appears twice?
The word APPLE can be arranged in 60 different ways
Explanation
The word APPLE has 5 letters
The condition is that the p should repeat twice
So, the number of arrangements without repetition is 5!
The number of arrangements with repetition is 2!
Then the arrangements = 5! / 2!
= 5 × 4 × 3 × 2! / 2!
= 5 × 4 × 3 = 60.
Problem 4
Six friends are to be seated around a round table. How many distinct seating arrangements are possible (considering rotations as identical)?
The number of arrangements is 120
Explanation
In the circular arrangement, the one position is fixed, the number of arrangements is (n - 1) and n is 6
So, (n - 1)! = (6 -1)! = 5!
5! = 5 × 4 × 3 × 2 ×1 = 120.
Problem 5
A family of 4 (father, mother, son, daughter) needs to be seated in a row for a photo, but the father must always sit at one of the ends. How many seating arrangements are possible?
The total arrangements are 12
Explanation
The father must sit at one of the ends which can be either left or right. Then the remaining 3 members.
So the number of ways they can be arranged is P(n, r) = n! / (n - r)!
Here, n and r = 3
P(3, 3) = 3! / (3 - 3)!
= 3! = 6
So the total number of arrangements = 2 × 6= 12.
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FAQs on Permutation and Combination
1. What is the difference between permutations and combinations?
2.What is the formula for permutations?
3.What is the formula for combination?
4.What is the relation between permutations and combinations?
5.How do we use permutations and combinations in real-life situations?
6.How can children in United States use numbers in everyday life to understand Permutation and Combination?
7.What are some fun ways kids in United States can practice Permutation and Combination with numbers?
8.What role do numbers and Permutation and Combination play in helping children in United States develop problem-solving skills?
9.How can families in United States create number-rich environments to improve Permutation and Combination skills?
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Jaipreet Kour Wazir
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Jaipreet Kour Wazir is a data wizard with over 5 years of expertise in simplifying complex data concepts. From crunching numbers to crafting insightful visualizations, she turns raw data into compelling stories. Her journey from analytics to education ref
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