Permutation and combination provide ways to represent a certain group of objects by choosing them in a set and creating subsets.

Maths permutations and combinations, in general, allow several methods to arrange a certain collection of data. Now comes the question: what is the definition of permutations and combination? When there is a selection of an object or data from a certain group of data then it is said to be a combination, whereas with the selection of data when the order of data comes into consideration then it is called permutations.

With this article on Permutation and combination learn more about permutation and combination formulas with solved examples, definitions, permutation vs combination table and more.

There is one more concept that is the repetition of data, both Permutation, and combination are divided into with repetition and without repetition and both concepts are very important in Mathematics. Will go through all such concepts in the article itself.

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Factorial

The continued product of first n natural numbers is denoted by n!

i.e

n! = 1 × 2 × 3 × ……. × n

Points to Remember

Key Points regarding factorial notation:

• 0! = 1
• Factorial of negative(-ve) figures and fractions is not specified.

\(\frac{n!}{r!}=n\left(n-1\right)\left(n-2\right)……..\left(r+1\right)\)

The Fundamental Principle of Multiplication

Let us assume there are two assignments A and B such that task A can be done in m different ways following which the second task B can be done in n different ways. Then the number of ways to accomplish the task A and B in succession respectively is given by m × n ways.

The Fundamental Principle of Addition

Let us assume there are two tasks A and B such that task A can be done in m different styles and assignment B can be completed in n ways. Then the number of ways to complete either of the two tasks is given by (m + n) ways.

Permutation Formula

Let us suppose there are 3 stools/chairs A, B, and C. So, the number of ways in which these chairs can be arranged is given by 3! = 6, wherein individual arrangement matters, and these arrangements are ABC, ACB, BCA, BAC, CAB, and CBA.

Mathematically, the permutation is associated with the act of arranging all the data of a set into some sequence or order.  Permutations come into account in more or less almost every domain of mathematics. The permutation formula is as follows:

• Consider r and n to be positive integers such that 0 ≤ r ≤ n. Then the number of ways to arrange “r” thing taken at a time out of n distinct things is given by:

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

• The number of methods to arrange n distinct things taken all at a time is given by:

\(^nP_n=n!\)

• The number of ways to arrange “r” things taken at a time out of n different things, wherein each thing may be repeated any number of times is given by:

\(n^r\)

Learn the various concepts of the Binomial Theorem here.

There are two types of permutation:

Repetition is Allowed

These are the simplest to determine.

Consider when a piece has n different types and one has r choices each time then the permutations is defined by:

n × n × … (r times)

This implies there are n possibilities for the first selection, followed by n possibilities for the second selection, and so on, multiplying each time.

No Repetition

In this case, each time the number of choices is reduced. Consider when a piece has n different types and one has r choices each time without repetition the permutations are:

n × (n-1) × (n-2)…

This implies there are n possibilities for the first selection, followed by n-1 possibilities for the second selection, and so on, multiplying each time.

Learn about Time and Work

Derivation of Permutation Formula

Let us consider that there are r cartons and each of them can carry one thing. There will be as many permutations as there are methods of filling in r empty carton by n things.

• Number of ways the first carton can be filled: n
• Number of ways the second carton can be filled: (n – 1)
• Number of ways the third carton can be filled: (n – 2)
• Number of ways the fourth carton can be filled: (n – 3)
• Number of ways \(r^{th}\) carton can be filled: [n – (r – 1)]
• The number of permutations of n distinct things taken r at a time, where 0 < r ≤ n and the objects do not replicate is:

n(n – 1)(n – 2)(n – 3) . . . (n – r + 1)

⇒ \(^nP_r = n(n – 1)(n – 2)(n – 3). . .(n – r + 1)\)

Multiplying and dividing by (n – r) (n – r – 1) . . . 3 × 2 × 1, we receive:

\(^nP_r=\frac{\left[n(n-1)(n-2)(n-3)…(n-r+1)(n-r)(n-r-1)…3\times2\times1\right]}{(n-r)(n-r-1)…3\times2\times1}=\frac{n!}{\left(n-r\right)!}\)

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

Check out this article on Number Systems.

Permutation and Combination Questions

Question1: How many 3-digit numbers can be formed utilizing digits 1, 2, 3, 4, 7, 9 lying within 300 and 500 if repetition of digits is permitted? 

Solution: Here, we have to form a 3 digit number lying between 300 and 500 using digits 1, 2, 3, 4, 7, 9 such that repetition of digits is allowed.

Unit’s digit can be filled by any one of the given numbers ⇒ No. of ways to fill unit’s digit = 6

Ten’s digit can be filled by any one of the given numbers ⇒ No. of ways to fill ten’s digit = 6

Hundreds of digits can only be filled by numbers 3, 4 ⇒ No. of ways to fill hundredth digit = 2

∴ The no. of ways to form a 3 digit number lying between 300 and 500 = 6 × 6 × 2 =72

Question 2: The number of words that can be created from all the characters of the word ‘CORONAVIRUS’? 

Solution: Here we have to create words using the letters of the word ‘CORONAVIRUS’

In total there are 11 letters in the word ‘CORONAVIRUS’ ⇒ n = 11.

The letter O is repeated twice in the word ‘CORONAVIRUS’ ⇒ r1 = 2.

The letter R is repeated twice in the word ‘CORONAVIRUS’ ⇒ r­2 = 2.

∴ The no. of words that can be formed using all the letters of the word ‘CORONAVIRUS’ =

\(\frac{11!}{2!\times2!}\)

Also, read about Relations and Functions here.

Circular Permutation

So far the type of permutation we studied is a straight permutation, the next is Circular Permutation. In this type of permutation, the number of ways to arrange n distinct things is given by: (n – 1)!

The number of ways to arrange n distinct things but all of them looks identical in a circular arrangement like the arrangement of beads in a necklace or the arrangement of flowers in a garland is given by: (n – 1)!/2

Now, let’s move towards the concept of combination;

Combination Formula

In combination, we have to select r things out of n given things wherein r ≤ n. So, the order in which we select these are things of the given n things does not matter. So, the number of ways to select r things out of n given things wherein r ≤ n is given by the below combination formula:

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

The combination is also classified as repetition allowed and repetition not allowed.

Key point:

The difference between permutation and combination is that permutation is all about the arrangement of things whereas combination is about selection where arrangement does not matter.

Learn more about probability with this article.

Derivation of Combination Formula

Let us consider that there are r cartons and all of them can carry one object.

• Number of ways to choose the first article/element from n distinct objects: n
• Number of ways to select the second object/element from (n-1) distinct objects: (n-1)
• Number of ways to select the third object/element from (n-2) distinct objects: (n-2)
• Number of ways to select \(r^{th}\) object/element from [n-(r-1)] different objects: [n-(r-1)]
• Completing the selection of r elements from the primary set of n things formulates an ordered subset of r elements.

∴ The number of ways to make a selection of r elements of the original or initial set of n elements is: n (n – 1) (n – 2) (n-3) . . . (n – (r – 1)) or n (n – 1) (n – 2) … (n – r + 1).

Let us suppose the ordered subset of r elements and all their permutations. Then the total estimate of all permutations of this subset is equivalent to r! because r objects in each combination can be ordered in r! ways.

Therefore, the total number of permutations of n distinct elements taken r at a time is \(^nC_r\times r!.\) It is nothing but:

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

\(^nC_r=\frac{nP_r}{r!}=\frac{n!}{r!\left(n-r\right)!}\)

Learn more about Lines of Regression here.

Relation Between Permutation and Combination

As we know, the number of ways to arrange things taken at a time out of n different things is given by:

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

Whereas, the number of ways to select r things out of n given things wherein r ≤ n is given by:

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

So from these two definitions, we can conclude that the relation between permutation and combination is as follows:

\(^nP_r=r!\times^nC_r\)

Permutation and Combination solved Problem

Question: Out of 7 consonants and 5 vowels, words are to be formed by involving 4 consonants and 3 vowels. The number of such words formed is?

Solution: Given that the No. of consonants = 7 and No. of vowels = 5.

Here we have to form words that contain 4 consonants and 3 vowels.

⇒ No. of ways to select 4 consonants out of 7 consonants =

\(^7c_4\)

⇒ No. of ways to select 3 vowels out of 5 vowels =

\(^5C_3\)

⇒ No. of words that can be formed containing 4 consonants and 3 vowels in it

=\(^7C_4\times^5C_3=350\)

Check more topics of Mathematics here.

Difference Between Permutation and Combination

The difference between permutations and combinations can be surmised by understanding the different conditions where the permutations and combinations theories are applied.

For grouping of articles/elements, or to arrange a count of the number of subgroups that can be reached from the provided set of things, we practice combinations. Moreover, to determine the number of possible arrangements of dissimilar elements, we practice permutations

Also, read about Arithmetic progressions with this article.

Permutation and Combination: Important Points

The following details would help us to compile the important points of the difference between permutation and combination.

• Permutations are used to obtain the count of the different number of arrangements that can be created with the given points.
• The combination is helpful to get the count of the number of distinct subgroups which can be developed from the provided larger set.
• For the identical values of n and r, the number of permutations or arrangements would always be larger than the number of combinations or groups.
• Seating arrangements for a related event or for official seating arrangements, the various possible seating arrangements can be determined by applying the formula of permutations.
• Furthermore, the different committees are created by selecting a few people, and the number of probabilities of building a committee is estimated by adopting the combinations formula.

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