Permutation and combination are possible ways of arranging and selecting elements from a group of elements without possible replacement. Both topics are very important in mathematics. They explain the various ways of arranging some sets of data.

Permutation and combination along with the differences between them will be properly explained in this article. In addition, keynotes with real-life examples will be discussed. As a bonus, students will watch a video lesson of simplified explanations of two questions on permutation and combinations to enhance their knowledge and heighten their boldness in solving future permutation and combination questions.


Permutation occurs almost in all facets of mathematics. Since it has to do with the arrangement of objects from a group of elements. So permutation is defined as the arrangement of objects or numbers in a precise order. The issue of permutation often arises when items like (balls, boxes, books, etc.) are to be arranged in a definite order. And it involves an arrangement in which the “order” is important.


We said earlier that permutation is the arrangement of elements in a specified order. So the arrangement of r items from a set of n objects in a precise order can be written as

{\color{DarkBlue} _{r}^{n}\textrm{P}=\frac{n!}{(n-r)!}}

Where \dpi{120} n= the number of items and \dpi{120} r= how many items are taken at a time.


In most cases, permutation includes the problems of arranging objects that are repeated. For example in the word AFRICA, we may be required to find the number of ways of arranging the letters of the word AFRICA. In this arrangement, the two A’s looks alike. Writing the two A’s with different kinds of letters such as AFRICa, will show that we have 6! ways of arranging the word.

The two A’s have 2! ways of arrangement without altering the other letters i.e. AFRICa or aFRICA. So the word AFRICA can be arranged in identical A’s 6!2! ways, i.e. \frac{6!}{2!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{2\times 1}

6!2!=6\times 5\times 4\times 3=360 ways

Furthermore, we can be given the word MINIMUM, and the asked to find the number of ways it can be arranged. In that word “MINIMUM” there are three M’s and two I’s, so the number of ways of arranging the 7 letter word such that three M’s and two I’s are identical is 7!3!2! ways.

So 7!3!2!=\frac{7!}{3!2!}=\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{3\times 2\times 1\times 2\times 1}

7!3!2!=\frac{7\times 6\times 5\times 4}{2}=\frac{840}{2}=420 ways

In general, the number of ways of arranging n objects of which r objects are identical is \dpi{120} {\color{DarkBlue} \frac{n!}{r!}}.  Where  

\dpi{120} n= numbers of ways of arranging all n objects if they are different and 

\dpi{120} r= number of ways of arranging the identical objects.


A combination is a mathematical technique of selecting objects or numbers from a group of objects or collections in such a way that the order of the selection does not matter. Note that combination involves “selection” in which the “order” of selection does not matter.


The combination is a selection of r elements from a set of n elements in which the order of selection does not matter. So the selection of r elements from a set of n elements can be written as

\dpi{120} _{r}^{n}\textrm{C}=\frac{_{r}^{n}\textrm{P}}{r!}=\frac{n!}{r!(n-r)!}

Where,  \dpi{120} n= number of elements and \dpi{120} r= elements taken at a time.

Permutation involves arrangementsCombination involves selections
Order is very important. For example, ab and ba are different arrangementsOrder is not important. For example, ab, and ba are the same selection or grouping
The formula for permutation is as aboveThe formula for Combination is as above
Table Showing the differences between Permutation and Combination

Note: The main difference between permutation and Combination is the ordering. With permutation, more attention is given to the order of the elements (objects or numbers). Whereas with combination, we don’t care much about the order. For Example, if you lock your locker or bag with a number key or lock “combo” with the number 7654, and try 4675 to open the locker, it won’t open. This is because it is a different number ordering, this is a permutation.

The equation connecting Combination and Permutation is \dpi{120} {\color{DarkBlue} _{r}^{n}\textrm{C}=\frac{_{r}^{n}\textrm{P}}{r!}}


A product such as 1\times 2\times 3 is called factorial 3, and it is written as 3! So factorial n i.e. n!=1\times 2\times 3\times .....\times (n-2)\times (n-1)\times n. Hence, the number of arrangements or permutation of n different object is _{n}^{n}\textrm{P}=n\times (n-1)\times (n-2)\times ...3\times 2\times 1-n! . And the number of arrangement of two objects taken from n different ones is _{2}^{n}\textrm{P}=n\times (n-1). Similarly _{3}^{n}\textrm{P}=n\times (n-1)(n-2) and _{4}^{n}\textrm{P}=n\times (n-1)(n-2)(n-3) etc. 

So the number of arrangements of r objects taken from n different ones is _{r}^{n}\textrm{P}=n(n-1)(n-2)....[n-(r-1)] or _{r}^{n}\textrm{P}=n(n-1)(n-2).... [n-r+1]. For example, _{4}^{6}\textrm{P}=6(6-1)(6-2)\times [6-(4+1)] =6\times 5\times 4\times (6-5)=6\times 5\times 4\times 3=360

Permutation and Combination Video Lessons 1

Permutation and Combination Video lesson 2

For more insights on these and more topics in Mathematics, download our e-books. These e-books are sure-best for your success, contain all topics in your syllabus plus past questions. Click HERE to download a copy Now!

Let us consider some practical examples on these topic and the solutions. This will consolidate all you have learnt in this mathematics article.


Example 1: In how many ways can four books be arranged in order if eight different books are available.


The first Shelf can be filled by one of the eight books 8 Ways
The Second Shelf is filled by one of the seven books left7 Ways
The third Shelf is filled by one of the six remaining books left6 Ways
The fourth Shelf is filled by one of the five remaining books left5 ways
The orderly arrangement of Books in the Shelf

Hence, the number of arrangement is \dpi{120} 8\times 7\times 6\times 5=1680

Example 2: If \dpi{120} _{r}^{6}\textrm{P}=6, find the value of \dpi{120} _{r+1}^{6}\textrm{P}

Solution:  \dpi{120} _{r}^{6}\textrm{P}=6 will give us \dpi{120} \frac{6!}{(6-r)!}=6

If \dpi{120} \frac{6!}{(6-r)!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{(6-r)!}

\dpi{120} \frac{6\times 5\times 4\times 3\times 2\times 1}{(6-r)!}=6

Cross multiplying the above expression will give 

\dpi{120} (6-r)!\times 6 = 6\times 5\times 4\times 3\times 2\times 1.

So \dpi{120} (6-r)!=\frac{6\times 5\times 4\times 3\times 2\times 1}{6}

\dpi{120} (6-r)!=5\times 4\times 3\times 2\times 1,

which can still be written as \dpi{120} (6-r)!=5!

\dpi{120} 6-r=5

\dpi{120} -r=5-6

\dpi{120} -r=-1,

\dpi{120} r=1.

Finally, \dpi{120} _{r+1}^{6}\textrm{P}= _{2}^{6}\textrm{P}= \frac{6!}{(6-2)!}=\frac{6!}{4!}

So \dpi{120} \frac{6!}{4!}=\frac{6\times 5\times 4\times 3\times 2\times 1}{4\times 3\times 2\times 1}=6\times 5=30

Example 3: A Committee of two men and three women is to be chosen from five men and four women. How many different committees can be formed?

Solution: The two men can be selected in \dpi{120} _{2}^{5}\textrm{C} ways.

Recall that generally, \dpi{120} _{r}^{n}\textrm{C}=\frac{_{r}^{n}\textrm{P}}{r!}

So \dpi{120} _{2}^{5}\textrm{C}=\frac{5\times 4}{2!}=\frac{20}{2\times 1}=10 ways

Note that  \dpi{120} _{2}^{5}\textrm{P}=\frac{5!}{(5-2)!}=\frac{5!}{3!}=\frac{5\times 4\times 3\times 2\times 1}{3\times 2\times 1}=5\times 4

Women can be chosen in \dpi{120} _{3}^{4}\textrm{C}=\frac{_{3}^{4}\textrm{P}}{3!}

\dpi{120} _{3}^{4}\textrm{C}=\frac{4\times 3\times 2}{3!}=\frac{4\times 3\times 2}{3\times 2\times 1}=4 ways

Also note that \dpi{120} _{3}^{4}\textrm{P}=\frac{4!}{(4-3)!}=\frac{4!}{1!}=4\times 3\times 2

So therefore, there are \dpi{120} 10\times 4=40 possible committees.

Conclusively, having seen the above examples, I want you to try the questions on permutation and combination below. If you do care to join our mathematics forum or community, were you will want to post your solutions to the questions, you are most welcomed. 


Try the three questions below. After which you may decide to check more questions on this topic from other sources.

  1. In how many ways can we arrange the word MATHEMATICS.
  2. In how many ways can the 1st 2nd and 3rd prizes be awarded in a race if there are 10 competitors.
  3. Find the number of ways of solving 8 subjects from 12 subjects for an examination.

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June 1, 2021

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