What is Permutation and Combination?
‘Permutation and combination’ is one of the most confusing topics in math. These calculation methods are used to select a definite number of items or data from a set or subset with or without restoring elements. In the case of permutations, elements are selected from a certain set.
It is a combination when the order in which items appear matters. In short, both permutation and combination discover how a certain data category can be organized.
Though these concepts are extremely important in mathematics, students often don’t get a command over this topic. Especially when it is about assignment writing on this complex topic, students couldn’t figure out the best way to proceed with their academic tasks.
Sometimes, students don’t have a clue about ‘What is Permutation and Combination?’. Therefore, doing their academic tasks on their own becomes immensely strenuous.
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However, we encourage you to write your assignments without any assistance. Hence, in order to help you clear all your concepts related to permutation and combination, this article comprises everything you need to get clarity on this topic. So, eliminate all your worries and dive into this article for valuable insights.
Let’s start with the definition of permutation and combination
If someone wants to be proficient with any math topic, it is crucial to understand its definition properly. Permutation is the technique to arrange the elements of a group or set in a particular progression.
Or, in case the objects or items are already arranged in a sequence when they are reorganized, it is known as permutation. It is among the most useful concepts in math and has its uses in almost every field.
Just like permutation, the combination also relates to the selection of elements from a group. But, when it comes to combination, the order of selecting the objects does not matter.
Moreover, if replacement is needed, it is known as combination replacement, and if replacement is not allowed, it is considered a combination only.
Now that you know what these terms mean, we can proceed further with the formula of permutations and combinations.
As explained above, the order of things is crucial in permutation.
Its primary formula is
P (n,r) = n! (n−r)!
Here,
In a set of n numbers of things, if you select r items at a particular time, the count of Permutations you will get is P (n,r). Another method to write it is nPr.
We can understand it like this:
Let us assume there is a giant bag with an ‘n’ number of objects. Also, there are an ‘ r’ number of empty bags, and you need to put one item in the different bags from the main bag that contains n elements.
You can just take an item to place in any one empty bag. Therefore, the different ways in which one can swap it is ‘n’
When it is about filling the second bag, you will be left with one less item to select from (as one element is already in the first bag). So, the probable ways to put things in the second bag are: ‘n-1’
In the same manner, the ways you can fill the third bag is ‘n-2’
The number of ways in which the rth number of bags, ‘t,’ can be filled is (n-(r-1)), where r is the total number of bags.
When you sum up all (the number of ways in which one can fill r bags one after the other), the total permutations you get is:
n (n – 1) (n – 2) (n-3) . . . (n – (r – 1))
Another way to write it is,
n (n – 1) (n – 2) … (n – r + 1)
Hence, if one takes r number of items at once out of the n different objects in a set, where 0 < r ≤ n (the total objects taken in the sample is either equal or lesser than the total items in the set though, it will always be greater than zero), then the total permutation one receives is:
⇒ nPr = n ( n – 1) ( n – 2)( n – 3). . .( n – r + 1)
Another way to write it is:
P (n,r) = n! (n−r)!
In Combination, the order of the articles you opt for does not matter.
Its primary formula is:
C (n,r) = n! (r! (n−r)!)
Here,
If a set has n number of things, and you have to choose r things out of it at a time, the total Combinations it can form is C (n,r). Another way to write it is nCr.
Let’s dive deeper into it:
Let’s consider there’s a big urn with ‘n’ different types of items. There are also empty urns that are ‘r’ in number. You need to keep one article in them, taking from the urn with ‘n’ objects.
You can take an object to put in the first empty urn. Therefore, you can fill it in n number of ways.
When it is about filling the second urn, you will have one item less to select from (as one item has already been taken out). So, the total ways in which you can fill the second urn is n-1
Likewise, the total ways in which the third urn can be filled is n-2
For the last urn or rth urn, the number of ways you can fill it is (n-(r-1)), where r is the actual number of urns.
When you do a sum of all these (the overall ways in which one can fill r urns in sequence), the number of combinations you finally get is:
n (n – 1) (n – 2) (n-3) . . . (n – (r – 1))
It can be rewritten as:
n (n – 1) (n – 2) … (n – r + 1)
With this, the difference between permutation combinations can be identified
When it is about permutation, you select items in a specific order. In the case of a combination, you can opt for items in any order. As a result, the total ways in which an article in a sample can be selected is ‘r’ (r is the total number of items in a sample)!
| Permutation | Combination |
| Organizing articles, items, numbers, people, objects, and alphabets | Selection of items, objects, clothes, students, and team. |
| Choosing a team captain or a pitcher from a group of people. | Selecting three members from a team. |
| Choosing two preferred words, in order, from a dictionary. | Selecting two words from a dictionary. |
| Choosing first, second, and third rankers. | Selecting three winners. |
A permutation is advantageous when there is a list of data (especially when the sequence of data is important) and the combination is of use for a collection of data (where the order of data is not considered).
Example 1: What will be the number of permutations and combinations of n = 6 and r = 2?
Solution:
Given, n = 6, r = 2
Using the formula for permutation:
nPr = (n!) / (n – r)!
⇒ nPr = (6!) / (6 – 2)!
⇒ nPr = 6! / 4! = (6× 5 × 4! )/ 4!
⇒ nPr = 30
For Combination:
nCr = n!/r!(n − r)!
⇒ nCr = 6!/2!(6 − 2)!
⇒ nCr = 6!/2!(4)!
⇒ nCr = 6 × 5 × 4!/2!(4)!
⇒ nCr = 15
Example 2: Find the number of ways a committee comprising 4 males and 2 females can be selected from 6 males and 5 females.
Solution:
Selecting 4 males out of 6 = 6C4 ways = 15 ways
Picking 2 females out of 5 = 5C2 ways = 10 ways
Members of the committee can be chosen as 6C4 × 5C2 = 150 ways.
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