Permutation Calculator

The permutations calculator calculates the number of ways you can arrange n distinct objects, taking a sample of r elements at a time. It tells us the number of possible arrangements of objects in groups where the order of arrangement is important. The total number of objects to be arranged is denoted by n, while the number of elements in each group is denoted by r.

For example, if we want to arrange the letters XYZ in groups of two letters each, then we will have XY, XZ, YZ, YX, ZX, and ZY: 6 ways.

To use this calculator, enter n, the total number of objects to be arranged in some order, and enter r, the number of elements in each group, and then click "Calculate".

Permutations

A permutation of a set is an arrangement of its members in a sequence or a particular order. If a set is already ordered, it is a permutation of its elements. For a permutation, the order of the elements is important. For example, permutations AB and BA are two different permutations. The number of permutations of n objects in samples of r objects is denoted as nPr.

The calculation of the number of permutations depends on the objects being arranged. It also depends on whether repetitions are allowed or not. Unless otherwise stated, we assume that repeats are not allowed when calculating permutations.

In this article we will look at examples of permutations without repetitions.

Permutations follow the fundamental principle of counting. It states that if an experiment consists of k events where the first event occurs n₁ times, the second event occurs n₂ events. So on till the event occurs nₖ times. The number of ways the experiment can occur sequentially is given by the product of the number of times the individual events occur, n₁ × n₂ × ... × nₖ.

Suppose we want to know the number of possible arrangements of the letters ABC with no repetitions in permutations. Any of the letters can come first, so there are 3 ways of setting the first letter.

After the first letter is set, there are two letters left, and either of the two letters can be set as the second letter, so there are two ways of setting the second letter. After the second letter is set, there will then be only one letter left. Thus, there is only one way to set the third letter.

Thus, by the fundamental counting principle, there are 3 × 2 × 1 = 6 ways to arrange the letters ABC. They are ABC, ACB, BCA, BAC, CAB, and CBA.

The Factorial

Above, we established that the number of permutations of 3 distinct objects is given by 3 × 2 × 1 = 6. Generally, the number of permutations of n objects (altogether) is given by n × (n-1) × (n-2) × ... × 1.

That is multiplications of all integers from n down to 1. The multiplication of all the integers from an integer, say n, down to 1 is called the factorial and is denoted by ! (the exclamation mark).

Thus, n!= n × (n-1) × (n-2) × ... × 1, and is called n factorial.

Note that 0!=1 and 1!=1.

The Example of Permutations

The standard track for races at the Olympics usually has 9 lanes. However, for the 100-meter race, lane 1 is usually not used. 8 runners are placed on lanes 2 to 9 in a row. How many possible ways can the 8 runners be arranged on lanes 2 to 9?

By the fundamental counting principle:

• any of the 8 runners gets lane 2,
• any of the remaining 7 runners can get lane 3,
• any of the remaining 6 runners can get lane 4,
• any of the remaining 5 runners can get lane 5,
• any of the remaining 4 runners can get lane 6,
• any of the remaining 3 runners can get lane 7,
• any of the remaining 2 runners may receive lane 8,
• one remaining runner receives lane 9.

Therefore, the total possible permutations of the 8 runners that can be arranged on the 8 tracks is 8! = 8 × 7 × 6 × ... × 2 × 1 = 40,320 ways.

In the permutations calculator, enter 8 in both the n (objects) and r (sample) boxes and click on Calculate to get 40,320.

Permutation of Subsets

In the previous examples, we looked at objects' permutations when all the objects are considered in the arrangements. However, there are situations when the objects are arranged into smaller groups.

In those cases, the total number of objects is donated by n, the number of objects in the groups (sample) is denoted by r, and the formula gives the number of permutations:

$$ₙPᵣ = \frac{n!}{\left(n-r\right)!}$$

This formula is used to calculate permutations without repeats. And if we need to organize in a certain order a sample r taken from the set n.

If we calculate the number of choices with which we can arrange all elements of the set in a certain order and without repetitions, we can use the following formula:

$$ₙPᵣ=n!$$

Example

In the above example, we looked at the number of possible ways all eight runners in a 100-meter race could be arranged. Now, in the same race, three medals are up for grabs. The first place in the race wins the gold medal, and the second and third-place runners win the silver and bronze medals, respectively. Out of the 8 runners in the race, how many possible ways can we get the gold, silver, and bronze medalists?

By the fundamental counting principle, any of the 8 runners can take the first position. After the first position had been filled, there would be seven runners left to contend for the second position. And after the second position, six runners would be in contention for the third position. Therefore, the total number of possible permutations of the first to third positions from the 8 runners is: 8 × 7 × 6 = 336

We use the formula:

$$ₙPᵣ = \frac{n!}{\left(n-r\right)!}$$

And we get

$$₈P₃=\frac{8!}{\left(8-3\right)!}=\frac{8!}{5!}=\frac{8×7×6×5×4×\ldots×1}{5×4×\ldots×1}=8×7×6=336$$

And in the permutations calculator, enter 8 in the n (objects) box and 3 in the r (sample) box and click on "Calculate" to get 336.

Permutations and Combinations: The Difference

Another essential counting technique is combinations. Combinations are the various ways a smaller number of objects (sample), r, can be selected from a larger number of objects, n. The number of combinations of r objects from n objects is denoted by simply ₙCᵣ.

In the definition of permutation, we mentioned that the order or arrangement is important. Well, that is the difference between permutations and combinations because, in combinations, the order is not important.

So, for example, we stated that the permutations of the letters XYZ in groups of two letters each will be the following XY, XZ, YZ, YX, ZX, and ZY. So we get six permutations.

However, the combinations of the letters XYZ in groups of two letters each are XY, XZ, and YZ; three combinations. This is because, in combinations, XY and YX are considered the same combinations; same with XZ and ZX, and same with YZ and ZY. Hence, the order of arrangement does not matter in calculating combinations.

The formula gives the number of combinations of r objects from n objects:

$$ₙСᵣ=\frac{n!}{r!\left(n-r\right)!}$$

Example of Сalculating Сombinations

In the above example with the runners, we obtained the number of ways we can select the first, second, and third positions from a group of 8 runners. Suppose we want to know the number of ways that 3 medalists can be selected from the group of 8 runners without considering their positions. It does not matter whether the person comes first, second, or third, as long as the runner wins a medal.

In this case, combinations are used because the order of the medals is unimportant. Thus, we solve this using the combinations formula.

$$ₙСᵣ = \frac{n!}{r!(n-r)!}$$

The number of ways 3 medalists can be selected from 8 runners is given by:

$$₈C₃=\frac{8!}{3! × \left(8-3\right)!}=\frac{8!}{3! × 5!}=\frac{8 × 7 × 6}{3 × 2 × 1}=8×7=56$$

Examples of calculating permutations

1. The news producer can choose 3 of the 5 guest speakers for their analytical program. The order of the guests is important. How many different ways can the producer arrange the guests' presentations? Order is important and repetition will not be used because the same guest cannot appear twice in the same news program. Therefore, we can use the formula for permutations.

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

$$₅P₃ = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 × 4 × 3 = 60$$

Thus we can see that the producer has 60 ways to organize the speakers.

2. A restaurant critic has selected 10 good establishments in town that serve sushi to rank the top 3 sushi restaurants. The establishments must be presented in an order that shows their place in the rankings. Also, the same place can not appear in the ranking several times. Thus, we satisfy the requirements for permutations formula - the order is important and there should be no repetitions. We use the formula for permutations:

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

$$₁₀P₃ = \frac{10!}{(10-3)!} = \frac{10!}{7!} = 10 × 9 × 8 = 720$$

3. When we say that order is important for permutations, it does not mean that the order must be numeric from 1 to, say, 10 or any other number. The order can be formed by certain objects between which we allocate our elements of the set.

For example, take the manager of an house repair company. He has four orders for painting rooms today. They are the office of a visa agency, a warehouse in a factory, a clothing store, and a room in a private home. The company has six painters. Each of them can go to 1 facility during one day. The remaining two painters will have the day off.

These objects are the office of a visa agency, a warehouse at a factory, a clothing store, and a room in a private home, which are analogs of positions 1, 2, 3, and 4.

The manager will have:

• 6 applicants who can be assigned to the office,
• 5 remaining applicants to be assigned to the warehouse,
• 4 remaining applicants to be sent to the store,
• 3 remaining applicants who can be assigned to a room in a private home.

So, intuitively, we can describe the number of choices as 6 × 5 × 4 × 3 = 360.

We are given the condition that the order in which the painters are distributed over the objects is important to us. No repetition is allowed, that is, a painter working on more than one object on the same day. So we can apply the permutation formula we already used.

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

$$\frac{6!}{(6-4)!} = \frac{6!}{2!} = 6 × 5 × 4 × 3 = 360$$

It turns out that there are 360 different ways that a home repair company manager can allocate orders among the available painters in given conditions.