Permutations and Combinations

Q: What is the difference between permutation and combination?

Permutation an arrangement or selection with order while combination is an arrangement or selection without order .

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When it comes to arrangement, it can be quite tricky. Imagine having a coach trying to make the selection for a football match from his team. In how many possible ways can he achieve this? Or perhaps a teacher looking for the best combination between girls and boys in his class to go for a road trip that requires just 7 students. In how many ways can he create his trip team of 7?

In this article, we will be learning about the ways that solve similar questions to the two mentioned above, while defining permutations and combinations.

What is permutation?

Permutation is the arrangement of objects by following a particular order or pattern. An ordered selection made is said to be permuted. We have two sorts of permutations: linear permutation and circular permutation.

Linear permutation

When an arrangement with order is done on a straight line, it is known as a linear permutation. For instance, when you are required to find the number of ways the following 6 balls are to be arranged in order. Such arrangement of objects done in a straight line can be seen in the image below.

Permutations and Combinations An image of linearly arranged balls for linear permutation StudySmarter

An image of linearly arranged balls for linear permutation, StudySmarter Originals

Circular permutation

However, when an arrangement with order is done in a circular or curved manner, it is known as a circular permutation. An example can be seen when different colored stones on a bead are to be arranged. The figure below gives an insight into circular arrangements.

Permutations and Combinations An illustration of balls arranged in a circular manner StudySmarter

An illustration of balls arranged in a circular manner, StudySmarter Originals

Unlike linear permutations where items are organized on a straight line, circular permutations are arranged in a circular manner as seen above. Hereafter, you shall have further details and examples regarding circular permutations.

What is the permutation sign?

Several signs are used to represent permutation; they are:

$P (n, r) o r {}_{n}P_{r}$

Types of linear permutation and corresponding formulas

There are two types of permutation, permutation with repletion and permutation with no repetition.

Permutation with repetition

In such a case of arrangement or selection, the object can be reused in all steps of selection. This means that each component of a group can be used whenever the selection is made.

For instance, if three letters from the alphabets A to Z were to be selected, then the letter A can be selected in those 3 times like AAA. The same argument goes for B and C and so on etc. Since we have 26 alphabets, we have 26 choices each time! Therefore, the amount of times the three letters would be selected is,

$26^{3} = 17576 t i m e s$

where the base 26 means that there are 26 alphabets and the exponent 3 represents the number of alphabets to be selected. This means that for selection with repetition using the formula,

$T h e n u m b e r o f p o s s i b l e w a y s = n^{r}$

where n is the number of objects in a set and r is the number of times the selection is made.

Permutation with no repetition

In this case, there is no repetition of a member of a set used when making selections. Once a member is used, it cannot be reused and as such reduces the chances available.

For instance, if one alphabet is to be picked and Z is picked, once Z is used it cannot be reused, reducing our chances from 26 to 25. Likewise, if two alphabets are to be picked and you pick Z and Y, once Z and Y are used, it further reduces our possibility from 26 to 24 and so on. Thus it leaves us with:

$26 \times 25 \times 24 \times 23 \times 22 \times 21 . . . \times 1 = 26! 26! = 403291461126605635584000000$

Therefore, unlike permutation with repetition which allows item among the sets to be replaced and reused, permutations without repetition does not allow the replacement and subsequent use of any item once used.

In what order can the letters A to F be arranged without repeating letters?

Solution:

In this event, a letter cannot appear more than once. So if A begins the set or takes the first position among the six vacant letter positions, then A must only remain in the first position without reappearing in another position unless it is removed from the first position and placed in another position. This rule applies to other letters in the set. Also, once A begins the sequence, it reduces the next letter by one chain. Such that A is the 6 chains of possibilities, B has 5 and C has 4 until F which has just a chain because it must have been repeated in all other chains.

Therefore the number of ways of arranging A to F orderly is:

$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 w a y s$

In the earlier example, all members of numbers are selected and arranged. Now, what happens if some are selected out of all? For instance, you are given numbers 1 to 10 and you have to select 6.

Recall we can arrange numbers 1 to 10 in

$10! = 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 w a y s$ .

However, since we are now selecting only 6 numbers, this means that we have

$10 \times 9 \times 8 \times 7 \times 6 \times 5 w a y s = \frac{10!}{4!} w a y s$

But $4! = (10 - 6)!$ . Thus we can deduce the permutation formula:

$P (n, r) = \frac{n!}{(n - r)!}$

where n is the number of objects, r is the number of objects to be selected from.

Therefore, to solve the question afresh, we have

$P (10, 6) = \frac{10!}{(10 - 6)!} P (10, 6) = \frac{10!}{4!} P (10, 6) = \frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1} P (10, 6) = 10 \times 9 \times 8 \times 7 \times 6 \times 5 P (10, 6) = 151, 200$

Regarding this formula, it suggests that if all elements in the set are to be selected and arranged, r becomes equal to n. Thus, it becomes expressed as,

$P (n, n) = \frac{n!}{(n - n)!} = \frac{n!}{0!} = n!$

since $0! = 1 .$

Using permutations in arranging letters

We recall that in order to make an ordered arrangement without repetition involving all members of a set of n members, we have P(n,n) = n! ways.

Meanwhile, if some members, r, out of a set, n, were to be selected and arranged, we have,

$P (n, r) = \frac{n!}{(n - r)!}$ ways

However, there are cases where a member of a set is repeated within a set. For instance, in the word WINNER, N is repeated twice. Therefore, to account for the double N it becomes:

$\frac{P (6, 6)}{P (2, 2)} = \frac{6!}{2!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 360$

Note that the permutation of all members is divided by the permutation of how many repetitions (in this case 2). So, if N were repeated thrice the permutation would have been divided by P(3,3) which is 3!.

Likewise, if you have more than one member of a set being repeated, the permutation of all members is divided by the product of the permutation of each repeated member. For instance, in the word LESSES, we have two E and three S, with a total of six alphabets. To arrange this we use:

$\frac{P (6, 6)}{P (2, 2) \times P (3, 3)} = \frac{6!}{2! \times 3!} = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (3 \times 2 \times 1)} = 60$

Based on this knowledge to arrange letters with the same alphabet it becomes

$\frac{n!}{p! q!}$

where, n is the total number of letters, p and q are the numbers of times an alphabet is repeated.

In how many ways can the word MISSISSIPPI be arranged?

Solution:

The word MISSISSIPPI contains 11 letters with I repeated 4 times, S repeated 4 times and P repeated 2 times.

Therefore, lets us apply the formula.

$\frac{n!}{p! q!}$

However, in this case, we have 3 repetitions so we do not stop at q, but add an s so our formula can be adjusted to:

$\frac{n!}{p! q! s!}$

Once this is applied we have:

$\frac{P (11, 11)}{P (4, 4) \times P (4, 4) \times P (2, 2)} = \frac{11!}{4! \times 4! \times 2!} \frac{11!}{4! \times 4! \times 2!} = \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) \times (2 \times 1)} \frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(4 \times 3 \times 2 \times 1) \times (4 \times 3 \times 2 \times 1) \times (2 \times 1)} = 34, 650$

Circular permutation formula

We have considered permutation of objects in a linear manner, but sometimes arrangement is done in a circular or rotating way as mentioned earlier in this study. This requires a different approach because unlike a straight line that begins from a point and ends at another point, a circle begins at a point and ends on the same point. This means that for an n set of numbers, n is repeated so that we have:

$\frac{P (n, n)}{n} = \frac{n!}{n} = \frac{n \times (n - 1)!}{n} \frac{n \times (n - 1)!}{n} = (n - 1)!$

Thus, permutation, in this case, follows the use of $(n - 1)!$ .

John has been asked to arrange 5 students at a round dining table. In how many ways can he achieve this?

Solution:

$n = 5 (n - 1)! = (5 - 1)! (5 - 1)! = 4! 4! = 4 \times 3 \times 2 \times 1 4 \times 3 \times 2 \times 1 = 24 w a y s$

What is combination?

The combination is a selection method that does not follow an order. Unlike permutation, if three letters were to be chosen from letters A to E, ABC, ACB, BAC, BCA, CAB, and CBA would all be outcomes because the order is involved. But in combination where the order is not needed, only one ABC stands for the rest because they are all repetitions if the order is not involved.

Outcomes of combination are lower than those of permutation because, with the removal of order, only one outcome replaces the orders which are similar. Like instead of writing these six outcomes; ABC, ACB, BAC, BCA, CAB, and CBA, you write just one ABC. reducing the option of that combination from 6 to 1.

To calculate combinations we use:

$C (n, r) = \frac{n!}{(n - r)! r!}$

Where

n stands for the total number of items to be chosen from

r stands for the number of items chosen.

What is the combination sign?

Several signs are used to represent permutation; they are:

$C (n, r) o r {}_{n}C_{r}$

Three students are to be chosen from a class of 8 to visit the museum, in how many ways can the decision be made?

Solution:

$n = 8 r = 3 C (n, r) = C (8, 3) C (8, 3) = \frac{8!}{(8 - 3)! \times 3!} \frac{8!}{(8 - 3)! \times 3!} = \frac{8 \times 7 \times 6 \times 5!}{5! \times 3!} \frac{8 \times 7 \times 6 \times 5!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3!}$

Note that:

$3! = 3 \times 2 \times 1 = 6 \frac{8 \times 7 \times 6}{3!} = 8 \times 7 = 56$

Therefore, three students can be chosen from a class of 8 to visit the museum in 56 ways.

Combination of multiple events

In combination with multiple events, more than one kind of choice is made from a set containing more than one group.

For instance, if a class of 14 contains 8 girls and 6 boys and you were to pick 4 boys and 5 girls, then you would have to consider your choice per gender. Therefore, it becomes picking 4 boys out of 6 and 5 girls out of 8. This means that:

$C (6, 4) \times C (8, 5) C (6, 4) = \frac{6!}{(6 - 4)! 4!} C (6, 4) = \frac{6 \times 5 \times 4!}{2! \times 4!} C (6, 4) = \frac{30}{2} C (6, 4) = 15$

and

$C (8, 5) = \frac{8!}{(8 - 5)! 5!} C (8, 5) = \frac{8 \times 7 \times 6 \times 5!}{3! \times 5!} C (8, 5) = \frac{8 \times 7 \times 6}{3 \times 2} C (8, 5) = 56$

therefore,

$C (6, 4) \times C (8, 5) = 15 \times 56 C (6, 4) \times C (8, 5) = 840$

Combination with repetition

Sometimes, selections are made without order but with repetition. Such operation is a combination with repetition and you apply the formula,

$\frac{(r + n - 1)!}{(n - 1)! r!}$

where n is the total number of things to choose from, r is the number of things we are to pick out of n and repetition is allowed with no order involved.

Dorothy has a collection of six differently colored billiard balls. If Kohe her friend is to select 4 balls out of these with color being repeated multiple times without order, in how many ways can this be achieved?

Solution:

Here, there is no order and the selection of a particular color is can be repeated. Thus, this is a combination with a repetition problem.

n (total number of balls) is 6

r (the number of balls to be selected) is 4

Thus, using:

$\frac{(r + n - 1)!}{(n - 1)! r!}$

The number of ways Kohe would achieve this is:

$Number ways = \frac{(4 + 6 - 1)!}{(6 - 1)! 4!} Number ways = \frac{9!}{5! 4!} Number ways = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! 4!} Number ways = \frac{9 \times 8 \times 7 \times 6}{4!} Number ways = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} Number ways = 126$

What is the difference between permutation and combination?

Confusing questions of permutation with those of combinations occur. They are indeed quite similar in that they both deal with possible items occurring. However, they differ in the following major ways:

Order

In permutation, the arrangement involves items that are arranged in order. This means that the position of items is very important. But in combination, the selection of items is without order. Therefore the order is not relevant in combination.

For example, if the letters A,B,C are to be permuted without repetition, then we have; AB, BA, AC, CA, BC, and CB. Meanwhile, if A, B, and C are to be combined without repetition, we have; AB, AC, and BC. Note that in permutation, AB and BA are not the same because, in AB, A comes before B same thing applies in BA and the rest of the outcomes. This as a matter of fact makes the outcomes of permutation larger than those of combination.

Terminology

Several people make mistakes when using the right terms regarding permutation as well as combination. In permutation, you arrange or organize but in combination, you choose or select. Never make the mistake of interchanging these terms when addressing either permutation or combination.

Formula

Very importantly and clearly observed, the formula of permutation differs from that of combination. the permutation formula is:

$P (n, r) = \frac{n!}{(n - r)!}$

however, the combination is calculated with:

$C (n, r) = \frac{n!}{(n - r)! r!}$

The difference in the formula is the r!, we can thus create a mathematical relationship between permutation and combination as:

$P e r m u t a t i o n = C o m b i n a t i o n \times r!$

This again confirms why the outcomes of permutation are larger than those of combination by a multiplicative factor of r!.

Further examples of permutation and combination

You should try out many more problems so as to have an idea of several ways you can be tasked in an exam. A few examples here would help you.

In how many ways can the letters of the word MALICE be written so that all consonant letters always stay next to each other?

Solution:

The word MALICE has 6 letters which include 3 consonants M, L and C. To arrange these consonants in the letters so that all consonant letters stay together means, an example of such arrangement would be MLCAIE. In order to do this, let us take all consonant letters to be just one letter since they appear as a group in several positions. This leaves us with 4 positions which include A, I, E and the group of consonants.

Next, find the number of ways arrangement can be done in all four positions.

$P (4, 1) = 4! = 4 \times 3 \times 2 \times 1 = 24$

So we can now find the number of ways these consonants can be arranged even as they stay close to each other.

$P (3, 1) = 3! = 3 \times 2 \times 1 = 6$

This means that the total number of ways the letters of the word MALICE can be arranged such that all consonants stay close to each other is

$24 \times 6 = 144$

A cutlery set contains 5 forks, 3 spoons and 4 knives. A boy picks 3 cutleries and he must pick at least a knife, in how many ways can he achieve this?

Solution:

If the boy must pick a knife there are three possible choices he can make:

1. Choose all 3 knives

In this case, ways he can achieve this is

$C (4, 3) = \frac{4!}{(4 - 3)! 3!} = \frac{4 \times 3!}{3!} = 4$

2. Choose 2 knives and any other 1 cutlery

In this case, ways he can achieve this is

$C (4, 2) \times C (8, 1) = \frac{4!}{(4 - 2)! 2!} \times \frac{8!}{(8 - 1)! 1!} = \frac{4 \times 3 \times 2}{2 \times 2} \times 8 = 48$

3. Choose 1 knife and any other 2 cutleries.

In this case, ways he can achieve this is

$C (4, 1) \times C (8, 2) = \frac{4!}{(4 - 1)! 1!} \times \frac{8!}{(8 - 2)! 2!} = 4 \times \frac{8 \times 7}{2} = 112$

So the number of ways he can select at least a knife among a selection of 3 cutleries is the sum of all possible events which is

$4 + 48 + 112 = 164$

Permutations and Combinations - Key takeaways

Permutation is the arrangement of objects or people by following an order or pattern.
Permutations could be with or without repetition.
Linear permutations are calculated using $\frac{n!}{(n - r)!}$ , while circular permutation is calculated using $(n - 1)!$ .
Combination is a selection method that does not follow an order.
Combination is calculated is using $\frac{n!}{(n - r)! r!}$
Permutation and combination differ in the importance and placement of the order, the terminology used and the formula applied.

Frequently Asked Questions about Permutations and Combinations

Permutation is the arrangement of objects or people by following an order or pattern.

Combination is a selection method that does not follow an order.

Permutation an arrangement or selection with order while combination is an arrangement or selection without order.

You use permutation when an arrangement or a selection is to be made with order, while the combination is used when an arrangement or selection is to be made without an order.

Permutation is calculated using n!/(n-r)!

Combination is calculated using n!/((n-r)!r!)

Flashcards in Permutations and Combinations14 Start learning

In what order can the letters V to Z be arranged?

120

In how many ways can two letters be picked from a selection of 5 letters?

In how many ways can 7 guests seat in a round table?

720

5 students are to be chosen from a set of 12 student. In how many ways can this be achieved?

792

There are 5 oranges marked A to E and 3 mangoes marked F to H. In how many ways can a selection of 2 mangoes and 4 oranges be made?

15 ways

In how many ways can the word SMILE be written so that all vowels are written close to each other?

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