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1274 LearnersLast updated on November 25, 2025

Mathematics has several concepts for counting or arranging 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.
Permutation and combination are ways to count things, and they help us solve many kinds of problems. A permutation is like arranging your toys on a shelf the order matters. If you put Teddy first, Robot second, and Car third, that’s different from putting Robot first and Teddy second.
A combination is like choosing a group of friends to play with the order doesn’t matter. Whether you say Sam, Mia, Leo, or Leo, Mia, Sam, it’s still the same group of three friends! Let’s explore more about permutations and combinations in a fun and straightforward way below.
The methods used to count the number of possible outcomes in a situation are permutations and combinations. 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:
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.
Permutation and combination formulas show us how many ways we can arrange items or make groups from a set of things.
Use a permutation when the order matters, like deciding who stands first, second, or third in a line.
Use a combination when the order doesn’t matter, like picking three friends to play a game together.
With the same number of items, there are always more permutations than combinations, because changing the order creates extra possibilities.
\(n! = n \times (n - 1) \times \cdots \times 3 \times 2 \times 1 \)
\(nCr = \frac{n!}{(n-r)! \, r!} \)
\( nPr = \frac{n!}{(n-r)!} \)


The counting problem is examined to determine whether to use permutations or combinations, and the corresponding formula is then applied accordingly.
Formula 1: Factorial of a number (n!)
A factorial is the product of all the whole numbers from 1 to n.
It’s like lining up numbers in a row and letting them all join hands!
\(n! = 1 \times 2 \times 3 \times 4 \times \cdots \times n \)
Formula 2: Permutations When Order Matters
This formula helps us find how many different arrangements we can make when we pick r things from n things, and the order matters.
Think of it like arranging friends in a line who stands first, second, or third makes a big difference!
\({}^nP_r = \frac{n!}{(n-r)!}, \quad \text{where } 0 \le r \le n \)
Formula 3: Permutations When Repetition Is Allowed
This formula is used when you can choose the same item multiple times. It’s like picking numbers for a secret code you can use the same number again and again!
Formula 4: Permutations When Some Items Repeat
This formula is used when you are arranging things, but some of them are the same.
It’s like arranging the letters in the word MOM the two M’s look the same, so swapping them doesn’t make a new arrangement.
\(\frac{n!}{p_1! \, p_2! \, \cdots \, p_k!} \)
Formula 5: Combinations When Order Doesn’t Matter
This formula helps us find how many different groups we can make when we pick r items from n items, and the order doesn’t matter.
\({}^nC_r = \frac{n!}{r!(n-r)!}, \quad \text{where } 0 \le r \le n \)
Permutations and combinations are used for counting arrangements and groups. Permutations are used when the order of items matters, as when we line up our toys or arrange participants in a race. Combinations are used when order does not matter, guiding us in choosing friends for a team or picking fruits for a delightful snack.
Permutation (Order Matters): Use permutation when the order of things is important. It’s like making a special line or sequence. For example:
Combination (Order Doesn’t Matter): Use combination when the order doesn’t matter, and you just want to make a group. For example:
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 |
The permutation formula is \(P(n,r) = {n! \over (n-r)!}\)
The combination formula is \(C(n,r) = {n! \over (n-r)! \space \times \space r!}\)
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.
Multiplying and dividing 1 with (n - r)!
⇒ \(P(n,r) = {{[n \times (n-1) \times ....... \times (n-r+1)] \times(n-r)!}\over (n-r)! }\)
Since, the factorial of (n - r) is:
\((n - r)! = (n - r) × (n - r - 1) × (n - r - 2) × ….. 3 × 2 × 1\)
⇒ \(P(n,r) = {{[n \times (n-1) \times ....... \times (n-r+1)] \times [(n-r) \times (n-r-1)....\times 2 \times1]}\over (n-r) \times (n-r-1)....\times 2 \times1 }\)
So, \(P(n,r) = {n! \over (n-r)!}\)
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, \(C(n,r) = {P(n,r) \over r!}\)
That is, \(C(n,r) = {n! \over (n-r)! \space \times \space r!}\)
Permutation and Combination Worksheet
Permutations and combinations are used to find the possible arrangements and selections of objects. Learning them improves analytical skills and helps solve counting problems easily.
Parents and teachers can help by:
Now, let’s learn a few common mistakes that students tend to make when working with 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.
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.
As all 5 books are different, we use permutation to find the number of arrangements.
\( P(n, r) = n! / (n - r)!\)
\(\text{Here, } r = 5 \text{ and } n = 5 \)
\(So, P(5, 5) = 5! / (5 - 5)!. \)
\(= 5! / 0! \)
\(5! = 5 × 4 × 3 × 2 × 1 = 120\)
\(0! = 1\)
\(So, 5! / 0! = 120.\)
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.
As the order is not considered when selecting the committee members, we use combinations.
\(C(n, r) = n! / (n - r)! r!\)
\(\text{Here, } n = 10 \text{ and } r = 4 \)
\(So, C(10, 4) =10! / (10 - 4)! 4!\)
\(= 10! / 4! × 6!\)
\(10! = 10 × 9 × 8 × 7 × 6! \)
Canceling 6! From numerator and denominator
\(= (10 × 9 × 8 × 7 × 6!) / (4! × 6!)\)
\(= 5040 / 4! \)
\( 4! = 4 × 3 × 2 × 1 = 24\)
\(So, 5040 / 24 = 210.\)
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.
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 \(= 5! / 2! \)
\(= 120 / 2 = 60.\)
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.
In the circular arrangement, the one position is fixed, the number of arrangements is (n - 1) and n is 6
\(So, (n - 1)! = (6 -1)! = 5! \)
\(5! = 5 × 4 × 3 × 2 ×1 = 120.\)
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.
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 \(= 2 × 6= 12.\)
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
: She compares datasets to puzzle games—the more you play with them, the clearer the picture becomes!






