What are Permutations and Combinations?

If there are 5 dishes, they can be relished in any order at a time. In permutation, it should be in a particular order. In combination, the order does not matter. Take 3 letters a, b, and c. The possible ways of pairing any two letters are ab, bc, ac, ba, cb and ca. It is in a particular order. So, this can be called the permutation of a, b, and c. But if the order does not matter then ab is the same as ba. Similarly, bc is the same as cb and ac is the same as ca. Here the list has ab, bc, and ac alone. This can be called the combination of a, b, and c.

How are Permutations and Combinations Helpful?

Imagine that you and your friends have planned to travel for the summer vacations by train to some country. You will have to plan and take all the necessary things in a suitcase. It is important to set a pin lock in the suitcase so that your things will be safe. If the pin is a three-digit number without repeating any number, how will you set it? You will choose a number from 0 to 9 and set it in the ones place. From 0 to 9 you have ten digits in total. You should not use this number anymore in the tens place and hundreds place. So, you will have to choose a number from the remaining 9 digits and set it in the tens place of the pin. How many more options are there for setting the last digit? You have already used two digits from 0 to 9, and you should not repeat those digits. So, you can choose a number from the remaining eight digits.

Now, if you forgot the last two digits and remember only the first digits, think of all the possible numbers which can occur in the last two digits. This is a very long process. Let us see some simple tricks to see how many possible ways the numbers can be without even listing the numbers.

Counting

Suppose you are having a grey shirt, a green shirt, and a red shirt and 2 pairs of trousers, one is blue and the other is black. Can you think of all possible pairs of shirts and trousers? One pair can be a gray shirt and a pair of blue trousers. Other can be a gray shirt with a pair of black trousers. Similarly, you can pair a green shirt with a pair of black trousers or with a pair of blue trousers. What about the red shirt? It can also be paired with a pair of blue trousers and a pair of black trousers. But how many pairs in total can you say? There are a total of 3 shirts and 2 pairs of trousers. 1 shirt can be paired with 2 pairs of trousers. There are 3 shirts, and they can be chosen in 3 different ways. There will be $3\times 2=6$ different pairs of shirts and trousers. This is how it is counted.

Practice Problem

If you have 5 cricket bats and 3 balls, in how many ways you can use them for playing?

Solution

There are a total of 5 cricket bats and 3 balls. 1 cricket bat can be used to play with 3 balls. There are 3 balls, and they can be chosen in 3 different ways. Therefore, the number of ways in which 5 cricket bats can be paired with 3 different balls will be $5\times 3=15$.

Counting Without Repetitions

In the word ring, there are four letters r, i, n, and g. Can you think of all possible four-letter words that can be formed by using these four letters? The words can also be meaningless, but letters should not be repeated. Put four boxes. If you put letter “i” in the first box, it should not occur in the other three boxes. For the remaining three boxes you can put r, n, or g. If you put r in the 2nd box, you cannot put r in the other 2 boxes. In the 3rd box, you can put n or g. If you put g in the 3rd box, then you have to fill the 4th box with the leftover letter n. So, for the 1st box, any one of the four letters can be used, in the 2nd box any one of the remaining three letters can be used, in the 3rd box, any one of the two remaining letters can be used and in the 4th box, the leftover letter is used. Example: ingr, rgni, gnir…The number of ways the four-letter words can be written without repeating the letters is $4\times 3\times 2\times 1=24$.

Counting With Repetitions

In the same arrangement of 4 letter words if it is allowed to repeat the letters, think about the 4 letter words that can be written. In the 1st box, any one of the 4 letters can be put. In the 2nd box, either use the same letter used in the 1st box or use the remaining 3 letters. Similarly, for the remaining 2 boxes, repeat the letters or use the remaining letters. Example: rrnn, rrrr, gnni, ginr…So, the number of ways the 4 letter words can be written with repeating letters is $4\times 4\times 4\times 4=256$.

Permutations

The previous example shows how many ways are there to write the four-letter word ring. Each arrangement of the word is unique and different from the other arrangement. Each arrangement is called the permutation of the four-letter word in which all the letters are taken at a time.

Let us now consider any three letters of the five-letter word ‘rings’. Without repeating the letters, list all possible words that can be framed. The words can even be meaningless. The words are inr, rni, nri, and so on in the list. By this, the counting of the permutation of five different letters in which three are taken at a time. The totals ways are $5\times 4\times 3=60$.

Here n=5, then n-1 is 5-1=4. Also, n-2 is 5-2=3. The total number of ways are $n\times \left(n-1\right)\times \left(n-2\right)$ in this case.

If the repetitions are allowed, the words are rrr, iii, iin, rin and so on in the list. The total ways are $5\times 5\times 5=125$.

Here n=5 and the letters can be repeated. The total ways can also be written as $n\times n\times n$.

The total permutation of n items taken r at a time is denoted as ${}^n{P}_r$.

Factorial

Instead of writing $n\times \left(n-1\right)\times \left(n-2\right)$, think of it in a simple way. Factorials will be helpful in this. It is denoted by the symbol ‘!’. While finding the factorial of 5, 1 is subtracted from 5(that is 5-1=4), 2 from 5(that is 5-2=3), 3 from 5 (that is 5-3=2) and keep doing that until it becomes 1. 5-4 will give 1. So, it should stop here. Then multiply all the numbers.

$5!=5\times \left(5-1\right)\times \left(5-2\right)\times \left(5-3\right)\times \left(5-4\right)$

$=5\times 4\times 3\times 2\times 1$

$=120$

Try finding 4!

$4!=4\times \left(4-1\right)\times \left(4-2\right)\times \left(4-3\right)$

$=4\times 3\times 2\times 1$

$=24$

The Formula for Finding Permutations

For the word ‘rings’, the permutation was taken for 5 different letters taken 3 at a time, without any repetitions.

$5!=5\times \left(5-1\right)\times \left(5-2\right)\times \left(5-3\right)\times \left(5-4\right)$ but the terms (5-3) and (5-4) are not needed when 3 letters are taken at a time. So, divide that term from 5! to get the permutation ${}^n{P}_r$where n=5 is the total number of letters and r is the number of letters taken at one time.

Here r=3.

Therefore,

$\begin{array}{c}{}^5{P}_3=\frac{5!}{\left(5-3\right)!}\\ =\frac{5\times 4\times 3\times 2\times 1}{2}\\ =60\end{array}$

Where,

$\begin{array}{c}\left(5-3\right)!=2!\\ =2\times \left(2-1\right)\\ =2\times 1\\ =2\end{array}$

This formula can be written in general as ${}^n{P}_r=\frac{n!}{\left(n-r\right)!}$.

Practice Problem

Find $\frac{4!}{5!}$.

Solution

Use the definition of $n!$ to calculate the values of $4!$ and $5!$.

$\frac{4!}{5!}=\frac{4\times (4-1)\times (4-2)\times (4-3)}{5\times (5-1)\times (5-2)\times (5-3)\times (5-4)}$

$=\frac{4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1}$

$=\frac{1}{5}$

Find $7!+3!$.

Solution

Use the definition of $n!$ to calculate the values of $7!$ and $3!$.

$7!+3!=7\times (7-1)\times (7-2)\times (7-3)\times (7-4)\times (7-5)\times (7-6)+3\times (3-1)\times (3-2)$

$=(7\times 6\times 5\times 4\times 3\times 2\times 1)+(3\times 2\times 1)$

$=5,040+6$

$=5,046$

Combinations

In combinations, the order does not matter. Of the three letters, ac is the same as ca if the order does not matter. It is similar to permutation except the order. The permutation of a, b, and c taken 2 at a time is

$\begin{array}{c}{}^3{P}_2=\frac{3!}{\left(3-2\right)!}\\ =\frac{3\times 2\times 1}{1}\\ =6\end{array}$

That is why it could be written it as 6 pairs ab, ba, bc, cb, ac, and ca. In the combinations, there are ab, ba, and ac. This is the same as permutation except that it should divide permutation ${}^n{P}_r$ by r!

The combination is denoted as ${}^n{C}_r$. The formula is

$\begin{array}{c}{}^n{C}_r=\frac{{}^n{P}_r}{r!}\\ =\frac{n!}{\left(n-r\right)!r!}\end{array}$

Here n=3 and r=2.

$\begin{array}{c}{}^3{C}_2=\frac{3!}{\left(3-2\right)!2!}\\ =\frac{3\times 2\times 1}{1\times 2\times 1}\\ =3\end{array}$

Find ${}^7{C}_2$.

Solution

$\begin{array}{c}{}^n{C}_r=\frac{{}^n{P}_r}{n!}\\ =\frac{n!}{(n-r)!r!}\end{array}$

Here, $n=7$ and $r=2$.

$\begin{array}{c}{}^7{C}_2=\frac{7!}{(7-2)!2!}\\ =\frac{7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\times 2\times 1}\\ =21\end{array}$

Common Mistakes

Do not confuse between permutations and combinations.

Formulas

The formula for the permutation of r objects from a total of n objects is:

${}^n{P}_r=\frac{n!}{\left(n-r\right)!}$

The formula for the combination of r objects from a total of n objects is:

$\begin{array}{c}{}^n{C}_r=\frac{{}^n{P}_r}{r!}\\ =\frac{n!}{\left(n-r\right)!r!}\end{array}$

Context and Applications

It is asked in K-12 and undergraduate entrance exams.

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

• Bachelors in Mathematics
• Masters in Mathematics