Permutation vs Combination: Definition and Formulas

What are Permutations and Combinations?

Permutation and combination are two fundamental concepts in mathematics that help us count and organize objects efficiently. They form the foundation of combinatorics, probability, statistics, and many real-world decision-making problems. While permutations focus on arrangements where the order matters, combinations deal with selections where the order does not matter. Understanding the difference between these concepts is essential for solving counting problems accurately.

Permutation Meaning in Simple Words

A permutation is an arrangement of objects in a specific order.

For example, if three students A, B, and C are standing in a line:

• ABC

• ACB

• BAC

• BCA

• CAB

• CBA

are all different permutations because the order changes.

In simple terms, a permutation answers the question:

"In how many ways can objects be arranged?"

Combination Meaning in Simple Words

A combination is a selection of objects where the order does not matter.

For example, if we select two students from A, B, and C:

• AB

• AC

• BC

These are combinations.

Notice that:

• AB and BA are considered the same combination.

In simple terms, a combination answers the question:

"In how many ways can objects be selected?"

Definition of Permutation

A permutation is an arrangement of objects in a particular order chosen from a set of objects.

For example: Arranging 3 books on a shelf.

Different arrangements create different permutations because position matters.

Definition of Combination

A combination is a selection of objects from a group without considering the order of selection.

For example: Selecting 3 players from a team of 10 players.

The order in which players are selected is irrelevant.

Why Permutations and Combinations are Important

Permutation and combination concepts are widely used in mathematics and practical applications.

Their importance includes:

• Solving probability problems.

• Data analysis and statistics.

• Computer science algorithms.

• Scheduling and planning.

• Cryptography and password security.

• Competitive examinations.

These topics frequently appear in JEE, CAT, CUET, SSC, Banking, NDA, UPSC, and university entrance examinations.

Basics of Permutation and Combination

Before learning formulas, it is important to understand the basic counting principles that form the foundation of permutations and combinations.

Fundamental Principle of Counting

The Fundamental Principle of Counting states that if one event can occur in $m$ ways and another independent event can occur in $n$ ways, then both events together can occur in:

$m\times n$ ways.

Example

Suppose:

• A shirt can be selected in 3 ways.

• A pair of trousers can be selected in 4 ways.

Total possible outfits: $3\times4=12$

This principle forms the basis of permutation and combination formulas.

Arrangements and Selections

Most counting problems can be classified into:

Arrangements

Order matters.

Examples:

• Arranging books.

• Seating students.

• Creating passwords.

Selections

Order does not matter.

Examples:

• Choosing a team.

• Selecting committee members.

• Picking lottery numbers.

Recognizing the difference is the first step toward choosing the correct formula.

When Does Order Matter?

Order matters when changing positions creates a different result.

Examples:

• Awarding Gold, Silver, and Bronze medals.

• Arranging letters in a word.

• Creating passwords.

For example:

ABC ≠ BAC

Since the arrangements are different, order matters.

These problems use permutations.

When Does Order Not Matter?

Order does not matter when only the selection is important.

Examples:

• Selecting a committee.

• Choosing friends for a trip.

• Picking lottery winners.

For example:

AB = BA

Both represent the same group.

These problems use combinations.

Permutation

Permutation focuses on arranging objects where the order is important.

What is a Permutation?

A permutation is an ordered arrangement of objects selected from a set.

Examples include:

• Arranging books.

• Seating guests.

• Assigning ranks.

Whenever position matters, permutations are used.

Permutation Formula

The standard permutation formula is: ${}^nP_r=\frac{n!}{(n-r)!}$

where:

• $n$ = total number of objects

• $r$ = number of objects selected

This formula calculates the number of ordered arrangements.

Types of Permutations

Permutations can be classified into different categories.

Linear Permutation

Objects are arranged in a straight line.

Example:

Arranging students in a row.

Permutation with Repetition

Objects may be repeated.

Example:

Creating passwords using digits.

Permutation without Repetition

Each object can be used only once.

Example:

Arranging different books.

Circular Permutations

In circular arrangements, objects are arranged around a circle.

Examples:

• People sitting around a round table.

• Circular race tracks.

Formula: $(n-1)!$

The reduction occurs because rotating the arrangement does not create a new permutation.

Combination

Combination focuses on selecting objects without considering order.

What is a Combination?

A combination is a selection of objects from a group where order does not matter.

Examples:

• Choosing team members.

• Selecting committee members.

• Picking lottery numbers.

Combination Formula

The standard combination formula is:

${}^nC_r=\frac{n!}{r!(n-r)!}$

where:

• $n$ = total objects

• $r$ = selected objects

This formula counts distinct selections.

Types of Combinations

Different situations require different combination methods.

Combination without Repetition

Each object can be selected only once.

Example:

Selecting students for a team.

Combination with Repetition

Objects may be selected multiple times.

Example:

Choosing flavors of ice cream.

Selection Without Arrangement

The defining feature of combinations is that arrangement is ignored.

For example:

Selecting A and B:

AB = BA

Thus only one combination exists.

Difference Between Permutation and Combination

Although both involve counting, they serve different purposes.

Order of Arrangement

Permutation: Order matters.

Combination: Order does not matter.

Formula Comparison

Permutation: ${}^nP_r=\frac{n!}{(n-r)!}$

Combination: ${}^nC_r=\frac{n!}{r!(n-r)!}$

Practical Interpretation

Permutation: "How many ways can objects be arranged?"

Combination: "How many ways can objects be selected?"

Comparison Table

Relationship Between Permutation and Combination

Permutations and combinations are closely connected mathematically.

Mathematical Relationship

The relationship is: ${}^nP_r=r!\times{}^nC_r$

This shows that permutations can be obtained from combinations by considering all possible arrangements.

Derivation of the Relationship

Starting with:

${}^nC_r=\frac{n!}{r!(n-r)!}$

Multiplying by $r!$:

$r!\times{}^nC_r=\frac{n!}{(n-r)!}$

which equals: ${}^nP_r$

Converting Permutations to Combinations

If a permutation value is known:

${}^nC_r=\frac{{}^nP_r}{r!}$

This conversion is frequently used in probability and combinatorics.

Important Observations

• Every permutation originates from a combination.

• Permutations are always greater than or equal to combinations.

• The difference is created by ordering.

Types of Permutations and Combinations

Different counting situations require different approaches.

Permutations Without Repetition

Each object is used only once.

Formula: ${}^nP_r=\frac{n!}{(n-r)!}$

Permutations With Repetition

Objects may repeat.

Formula: $n^r$

Combinations Without Repetition

Each object is selected only once.

Formula: ${}^nC_r=\frac{n!}{r!(n-r)!}$

Combinations With Repetition

Formula: ${}^{n+r-1}C_r$

Used when selections can repeat.

Properties of Permutations and Combinations

Several mathematical properties simplify calculations.

Factorial Properties

Factorials are fundamental to counting.

Examples:

• $0!=1$

• $1!=1$

• $5!=120$

Symmetry Property of Combinations

A useful property is:

${}^nC_r={}^nC_{n-r}$

Example:

${}^{10}C_3={}^{10}C_7$

Recursive Relationships

Combination values satisfy:

${}^nC_r={}^{n-1}C_r+{}^{n-1}C_{r-1}$

This forms the basis of Pascal's Triangle.

Important Counting Properties

• ${}^nC_0=1$

• ${}^nC_n=1$

• ${}^nP_n=n!$

These properties are frequently used in exam questions.

Applications of Permutations and Combinations

Permutation and combination concepts appear across many disciplines.

Applications in Probability

Used for:

• Card games.

• Dice problems.

• Lottery calculations.

• Event probabilities.

Applications in Statistics

Used for:

• Sampling.

• Data analysis.

• Experimental design.

Applications in Computer Science

Used for:

• Algorithm design.

• Password generation.

• Cryptography.

• Data structures.

Applications in Daily Life

Examples include:

• Creating schedules.

• Seating arrangements.

• Team selection.

• Planning events.

Permutation vs Combination Formula Chart

This section provides a quick reference for important formulas.

Permutation Formula Table

Combination Formula Table

Factorial Formula Reference

Quick Comparison Chart

Common Mistakes in Permutation and Combination

Students often lose marks because they choose the wrong counting method.

Confusing Arrangement with Selection

Always determine whether the question involves arranging or selecting objects.

Ignoring Order Conditions

The most common mistake is forgetting to check whether order matters.

Incorrect Use of Factorials

Common errors include:

• Forgetting $0!=1$

• Cancelling factorials incorrectly

• Using wrong factorial values

Choosing the Wrong Formula

Before solving any problem, ask:

"Does order matter?"

If yes, use permutations.

If no, use combinations.

This single question can prevent most mistakes in permutation and combination problems.

Best Books for Permutation and Combination

Permutation and combination form the foundation of counting principles, probability, and combinatorics. The following books provide detailed explanations and extensive practice questions.

Shortcut Tips and Tricks for Permutation and Combination

Many counting problems can be solved quickly by identifying whether order matters or not. These shortcuts help avoid common mistakes.

Important Formula Table

These are the most commonly used formulas in permutation, combination, and counting principles.

Solved examples Based on Permutations Vs Combinations

Example 1: If ${}^{2n}C_3 : {}^nC_3 = 10 : 1$, then find the ratio $(n^2+3n):(n^2-3n+4)$. (JEE Main 2023)

Solution:

$\frac{{}^{2n}C_3}{{}^nC_3}=10$

$\Rightarrow \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}}=10$

$\Rightarrow \frac{(2n)!(n-3)!}{(2n-3)!n!}=10$

$\Rightarrow \frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)}=10$

$\Rightarrow \frac{4(2n-1)}{n-2}=10$

$\Rightarrow 8n-4=10n-20$

$\Rightarrow 2n=16$

$\Rightarrow n=8$

Therefore,

$\frac{n^2+3n}{n^2-3n+4}=\frac{8^2+3(8)}{8^2-3(8)+4}$

$=\frac{64+24}{64-24+4}$

$=\frac{88}{44}$

$=2$

Hence, the required ratio is $2:1$.

Example 2: The number of different ways in which five alike dashes and eight alike dots can be arranged using only seven of these dashes and dots is:

Solution:

Since there are only 5 dashes available, the number of dashes selected can be:

$0,1,2,3,4,5$

The remaining symbols will be dots.

If $r$ dashes are selected, then the number of arrangements of 7 symbols is

${}^7C_r$

Therefore, the total number of arrangements is

${}^7C_0+{}^7C_1+{}^7C_2+{}^7C_3+{}^7C_4+{}^7C_5$

Using the identity

$\sum_{r=0}^{7}{}^7C_r=2^7$

we get

$=2^7-\left({}^7C_6+{}^7C_7\right)$

$=128-(7+1)$

$=120$

Hence, the answer is $120$.

Example 3: All the five-digit numbers $N=abcde$ having the property $a<b<c<d<e$ are arranged in increasing order of magnitude. The $97^{\text{th}}$ number in the list does not contain the digit:

Solution:

Every such number is formed by selecting 5 distinct digits from ${1,2,3,4,5,6,7,8,9}$.

Since the digits must be in increasing order, each selection gives exactly one number.

Total numbers:

${}^9C_5=126$

Numbers beginning with 1:

${}^8C_4=70$

(using digits from ${2,3,4,5,6,7,8,9}$)

Thus, the first 70 numbers start with 1.

Now consider numbers beginning with 23.

Remaining digits are chosen from ${4,5,6,7,8,9}$.

Number of such numbers:

${}^6C_3=20$

Therefore,

$70+20=90$

The first 90 numbers are covered.

Now consider numbers beginning with 245.

The remaining two digits are chosen from ${6,7,8,9}$.

Number of such numbers:

${}^4C_2=6$

These numbers are:

24567

24568

24569

24578

24579

24589

Thus,

$90+6=96$

Therefore, the $97^{\text{th}}$ number is

$24678$

The digits present are $2,4,6,7,8$.

Hence, the digit not present is $5$.

Example 4: A seven-digit number is in the form $abcdefg$, where $a<b<ce>f>g$ and $a,b,c,e,f,g$ are distinct digits. Find the number of such numbers.

Solution:

Case 1: Zero is not used

Select 7 digits from ${1,2,3,4,5,6,7,8,9}$.

Number of selections:

${}^9C_7$

Among the selected digits, the largest digit must occupy position $d$.

From the remaining 6 digits, choose any 3 digits for positions $a,b,c$.

Number of ways:

${}^6C_3$

Thus, total numbers in this case are

${}^9C_7\times{}^6C_3$

Case 2: Zero is used

Since the first digit cannot be zero and the sequence after $d$ is decreasing, zero must occupy the last position.

Choose the remaining 6 digits from ${1,2,3,4,5,6,7,8,9}$.

Number of selections:

${}^9C_6$

Again, the largest selected digit occupies position $d$.

Choose 3 of the remaining 5 digits for positions $a,b,c$.

Number of ways:

${}^5C_3$

Thus, total numbers in this case are

${}^9C_6\times{}^5C_3$

Hence, total numbers

$={}^{9}C_7\times{}^6C_3+{}^9C_6\times{}^5C_3$

$={}^{9}C_2\times{}^6C_3+{}^9C_3\times{}^5C_3$

$=36\times20+84\times10$

$=720+840$

$=1560$

Hence, the answer is $1560$.

Example 5: A 6-digit number is to be formed using the digits $0-9$, where repetition is allowed. How many different numbers can be formed if the number must be divisible by 2 and have exactly 3 even digits?

Solution:

Since the number is divisible by 2, the last digit must be even.

There are 5 choices for the last digit:

${0,2,4,6,8}$

Now among the remaining 5 positions, exactly 2 positions must contain even digits and 3 positions must contain odd digits.

Choose the positions of the 2 even digits:

${}^5C_2=10$

Each selected even position can be filled in 5 ways.

Therefore,

$5^2=25$

The remaining 3 positions must contain odd digits.

Each odd position can be filled in 5 ways.

Therefore,

$5^3=125$

Total arrangements:

$={}^{5}C_2\times5^2\times5^3\times5$

$=10\times25\times125\times5$

$=156250$

Hence, the number of such 6-digit numbers is

$156250$.

Related Topics to Permutation and Combination

Permutation and combination are important topics in combinatorics and probability. Studying related counting techniques and probability concepts can help develop a deeper understanding of selection, arrangement, and mathematical reasoning problems.