Permutations and Combinations

Permutation and Combination are the most fundamental concepts in mathematics related to picking items from a group or set.

• Permutation is arranging items considering the order of selection from a certain group.
• Combination is selecting items without considering order.
• For example, in the below diagram, PQ and QP are different in permutation but same in combination. Therefore we have more permutations than combinations.

Permutation Meaning

Permutation is the distinct interpretations of a provided number of components carried one by one, or some, or all at a time. For example, if we have two components A and B, then there are two likely performances, AB and BA.

A numeral of permutations when ‘r’ components are positioned out of a total of ‘n’ components is ⁿ P_r. For example, let n = 3 (A, B, and C) and r = 2 (All permutations of size 2). Then there are ³P₂ such permutations, which is equal to 6. These six permutations are AB, AC, BA, BC, CA, and CB. The six permutations of A, B, and C taken three at a time are shown in the image added below:

Permutation Formula

Permutation formula is used to find the number of ways to pick r things out of n different things in a specific order, and replacement is not allowed, and is given as follows:

Explanation of Permutation Formula

As we know, permutation is an arrangement of r things out of n where the order of arrangement is important( AB and BA are two different permutations). If there are three different numerals 1, 2 and 3 and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That it can be accomplished in 6 methods. 

Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways. 

In general, n distinct things can be set taking r (r < n) at a time in n(n - 1)(n - 2)...(n - r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n - 1 things. Likewise, the third thing can be any of the remaining n - 2 things. Alike, the r^th thing can be any of the remaining n - (r - 1) things. 

Hence, the entire number of permutations of n distinct things carrying r at a time is n(n - 1)(n - 2)...[n - (r - 1)], which is written as ⁿ P_r. Or, in other words, 

\bold{{}^nP_r = \frac{n!}{(n-r)!} }

Combination Meaning

It is the distinct sections of a shared number of components carried one by one, or some, or all at a time. For example, if there are two components A and B, then there is only one way to select two things, select both of them.

For example, let n = 3 (A, B, and C) and r = 2 (All combinations of size 2). Then there are ³C₂ such combinations, which is equal to 3. These three combinations are AB, AC, and BC.

Here, the combination of any two letters out of three letters A, B, and C is shown below, we notice that in combination the order in which A and B are taken is not important as AB and BA represent the same combination.

Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items i.e., two distinct permutation, but for selecting, AB and BA are the same i.e., same combination.

Combination Formula

Combination Formula is used to choose ‘r’ components out of a total number of ‘n’ components, and is given by:

Using the above formula for r and (n-r), we get the same result. Thus,

\bold{{}^nC_r = {}^nC_{(n-r)}}

Explanation of Combination Formula

Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3). 

Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as ³C₂. In general, the number of combinations of n distinct things taken r at a time is, 

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

Derivation of Permutation and Combination Formulas

We can derive these Permutation and Combination formulas using the basic counting methods, as these formulas represent the same thing. Derivation of these formulas is as follows:

Derivation of Permutations Formula

Permutation is selecting r distinct objects from n objects without replacement, and where the order of selection is important. By the fundamental theorem of counting and the definition of permutation, we get

P (n, r) = n . (n-1) . (n-2) . (n-3).  . . .  .(n-(r+1))

By multiplying and dividing above with (n-r)! = (n-r).(n-r-1).(n-r-2).  . . .  .3. 2. 1, we get

P (n, r) = [n.(n−1).(n−2)….(nr+1)[(n−r)(n−r−1)(n-r)!] / (n-r)!

⇒ P (n, r) = n!/(n−r)!

Thus, the formula for P (n, r) is derived.

Derivation of Combinations Formula

Combination is choosing r items out of n items when the order of selection is of no importance. Its formula is calculated as,

C(n, r) = Total Number of Permutations /Number of ways to arrange r different objects. 
[Since by the fundamental theorem of counting, we know that number of ways to arrange r different objects in r ways = r!]

C(n,r) = P (n, r)/ r!

⇒ C(n,r) = n!/(n−r)!r!

Thus, the formula for Combination i.e., C(n, r) is derived.

Read in Detail - [Combinations Formulas in Maths]

Permutation Vs Combination

Differences between permutation and combination can be understood by the following table:

Also Check,

• Binomial Theorem
• Binomial Expansion
• Binomial Random Variables
• Fundamental Theorem of Counting

Solved Examples on Permutation and Combination

Example 1: Find the number of permutations and combinations of n = 9 and r = 3.

Solution: 

Given, n = 9, r = 3

Using the formula given above:

For Permutation:

ⁿP_r = (n!) / (n - r)! 
⇒ ⁿP_r = (9!) / (9 - 3)! 
⇒ ⁿP_r = 9! / 6! = (9 × 8 × 7 × 6! )/ 6! 
⇒ ⁿP_r = 504

For Combination:

ⁿC_r = n!/r!(n − r)!
⇒ ⁿC_r = 9!/3!(9 − 3)!
⇒ ⁿC_r = 9!/3!(6)!
⇒ ⁿC_r = 9 × 8 × 7 × 6!/3!(6)!

⇒ ⁿC_r = 84

Example 2: In how many ways can a committee consisting of 4 men and 2 women can be chosen from 6 men and 5 women?

Solution:

• Choose 4 men out of 6 men = ⁶C₄ ways = 15 ways
• Choose 2 women out of 5 women = ⁵C₂ ways = 10 ways

The committee can be chosen in ⁶C₄ × ⁵C₂  = 150 ways.

Example 3: In how many ways can 5 different books be arranged on a shelf?

Solution: 

This is a permutation problem because the order of the books matters. 

Using the permutation formula, we get:
⁵P₅ = 5! / (5 - 5)! = 5! / 0! = 5 x 4 x 3 x 2 x 1 = 120

Therefore, there are 120 ways to arrange 5 different books on a shelf.

Example 4: How many 3-letter words can be formed using the letters from the word "FABLE"?

Solution: 

This is a permutation problem because the order of the letters matters. 

Using the permutation formula, we get:
⁵P₃ = 5! / (5 - 3)! = 5! / 2! = 5 x 4 x 3 = 60

Therefore, there are 60 3-letter words that can be formed using the letters from the word "FABLE".

Example 5: A committee of 5 members is to be formed from a group of 10 people. In how many ways can this be done?

Solution: 

This is a combination problem because the order of the members doesn't matter. 

Using the combination formula, we get:

¹⁰C₅ = 10! / (5! x (10 - 5)!) = 10! / (5! x 5!) 
⇒¹⁰C₅= (10 x 9 x 8 x 7 x 6) / (5 x 4 x 3 x 2 x 1) = 252

Therefore, there are 252 ways to form a committee of 5 members from a group of 10 people.

Example 6: A pizza restaurant offers 4 different toppings for their pizzas. If a customer wants to order a pizza with exactly 2 toppings, in how many ways can this be done?

Solution: 

This is a combination problem because the order of the toppings doesn't matter. 

Using the combination formula, we get:
⁴C₂ = 4! / (2! x (4 - 2)!) = 4! / (2! x 2!) = (4 x 3) / (2 x 1) = 6

Therefore, there are 6 ways to order a pizza with exactly 2 toppings from 4 different toppings.

Example 7: How many considerable words can be created by using 2 letters from the term“LOVE”?

Solution: 

The term “LOVE” has 4 distinct letters.

Therefore, required number of words = ⁴P₂ = 4! / (4 – 2)!

Required number of words = 4! / 2! = 24 / 2

⇒ Required number of words = 12

Example 8: Out of 5 consonants and 3 vowels, how many words of 3 consonants and 2 vowels can be formed?

Solution:

Number of ways of choosing 3 consonants from 5 = ⁵C₃
Number of ways of choosing 2 vowels from 3 = ³C₂
Number of ways of choosing 3 consonants from 2 and 2 vowels from 3 = ⁵C₃ × ³C₂

⇒ Required number = 10 × 3
= 30

It means we can have 30 groups where each group contains a total of 5 letters (3 consonants and 2 vowels).

Number of ways of arranging 5 letters among themselves
= 5! = 5 × 4 × 3 × 2 × 1 = 120

Hence, the required number of ways = 30 × 120

⇒ Required number of ways = 3600

Example 9: How many different combinations do you get if you have 5 items and choose 4?

Solution:

Insert the given numbers into the combinations equation and solve. “n” is the number of items that are in the set (5 in this example); “r” is the number of items you’re choosing (4 in this example):

C(n, r) = n! / r! (n – r)! 
⇒ ⁿC_r = 5! / 4! (5 – 4)!
⇒ ⁿC_r = (5 × 4 × 3 × 2 × 1) / (4 × 3 × 2 × 1 × 1)
⇒ ⁿC_r = 120/24 
⇒ ⁿC_r = 5

The solution is 5.

Example 10: Out of 6 consonants and 3 vowels, how many expressions of 2 consonants and 1 vowel can be created?

Solution:

Number of ways of selecting 2 consonants from 6 = ⁶C₂
Number of ways of selecting 1 vowels from 3 = ³C₁
Number of ways of selecting 3 consonants from 7 and 2 vowels from 4.
⇒ Required ways = ⁶C₂ × ³C₁
⇒ Required ways = 15 × 3
⇒ Required ways= 45

It means we can have 45 groups where each group contains a total of 3 letters (2 consonants and 1 vowels).

Number of ways of arranging 3 letters among themselves = 3! = 3 × 2 × 1
⇒ Required ways to arrenge three letters = 6

Hence, the required number of ways = 45 × 6

⇒ Required ways = 270

Example 11: In how many distinct forms can the letters of the term 'PHONE' be organized so that the vowels consistently come together?

Solution:

The word 'PHONE' has 5 letters. It has the vowels 'O',' E', in it and these 2 vowels should consistently come jointly. Thus these two vowels can be grouped and viewed as a single letter. That is, PHN(OE).

Therefore we can take total letters like 4 and all these letters are distinct.

Number of methods to organize these letters = 4! = 4 × 3 × 2 × 1
⇒ Required ways arrenge letters = 24

All the 2 vowels (OE) are distinct.

Number of ways to arrange these vowels among themselves = 2! = 2 × 1
⇒ Required ways to arrange vowels = 2

Hence, the required number of ways = 24 × 2
⇒ Required ways = 48.

Practice Questions - Permutations and Combinations

Question 1: How many ways can 5 books be arranged on a shelf?

Question 2: In how many ways can a committee of 3 be chosen from a group of 10 people?

Question 3: How many 4-digit PIN codes can be created using the digits 0-9 if repetition is allowed?

Question 4: A pizza shop offers 8 toppings. How many different 3-topping pizzas can be made?

Question 5: In how many ways can 6 people be seated at a round table?

Question 6: How many ways are there to select 2 boys and 3 girls from a class of 5 boys and 6 girls?

Question 7: How many different 5-card hands can be dealt from a standard 52-card deck?

Question 8: In how many ways can the letters of the word "MATHEMATICS" be arranged?

Question 9: A bag contains 5 red marbles, 3 blue marbles, and 2 green marbles. How many ways are there to select 4 marbles?

Question 10: How many ways are there to arrange the letters in the word "MISSISSIPPI"?

Summary

Permutations and combinations are fundamental concepts in probability and statistics used to calculate the number of possible outcomes in various scenarios. Permutations deal with arrangements where order matters, calculated using the formula P(n,r) = n! / (n-r)!, where n is the total number of items and r is the number being arranged. Combinations, on the other hand, focus on selections where order is irrelevant, using the formula C(n,r) = n! / (r! * (n-r)!). The key difference lies in whether the order of selection is important. Factorials, denoted as n!, play a crucial role in both calculations, representing the product of all positive integers up to and including n. These concepts find applications in diverse fields, from creating PIN codes and arranging seating plans (permutations) to selecting team members and calculating lottery odds (combinations). Understanding when to apply each concept is essential for solving problems involving counting and probability in both academic and real-world contexts.