Permutation Calculator

Our permutations calculator determines the exact number of ways you can arrange n distinct objects, taking a sample of r elements at a time. It calculates the number of possible arrangements for groups where the specific sequence or order of elements is strictly important. The total number of available objects is denoted by n, while the number of elements in each selected group is denoted by r.

For example, if we want to arrange the letters XYZ into groups of two letters each, we can form XY, XZ, YZ, YX, ZX, and ZY, resulting in 6 distinct ways.

To use this nPr calculator, simply enter n (the total number of objects to be arranged) and r (the number of elements in each sample group), then click "Calculate".

Permutations

In mathematics, a permutation is an arrangement of a set's members into a specific sequence or order. If a set is already ordered, rearranging its elements creates a new permutation. In any permutation, the order of the elements absolutely matters. For example, the sequences AB and BA represent two entirely different permutations. The total number of permutations of n objects taken in samples of r objects is commonly denoted as nPr.

Calculating the number of permutations depends heavily on the type of objects being arranged and whether repetitions are permitted. Unless otherwise stated, it is generally assumed that repetitions are not allowed when calculating permutations.

In this article, we will focus exclusively on examples of permutations without repetition.

Permutations rely on the fundamental principle of counting. This principle states that if an experiment consists of k sequential events, where the first event can occur in n₁ ways, the second in n₂ ways, and so on until the final event occurs in nₖ ways, the total number of ways the experiment can unfold is the product of these individual occurrences: n₁ × n₂ × ... × nₖ.

Suppose we want to determine the number of possible permutations for the letters ABC without any repetitions. Any of the three letters can be placed first, meaning there are 3 ways to set the first letter.

Once the first letter is set, two letters remain. Either of these two can be chosen as the second letter, giving us 2 ways to set the second letter. After the second letter is selected, only one letter remains, meaning there is only 1 way to set the third letter.

By applying the fundamental counting principle, there are 3 × 2 × 1 = 6 total ways to arrange the letters ABC. These arrangements are ABC, ACB, BCA, BAC, CAB, and CBA.

The Factorial

As shown above, the number of permutations of 3 distinct objects is calculated as 3 × 2 × 1 = 6. In general, the number of permutations for arranging an entire set of n objects is given by n × (n-1) × (n-2) × ... × 1.

This involves multiplying all positive integers from n down to 1. In mathematics, the product of an integer n and all the positive integers below it is called a factorial, which is denoted by the exclamation mark (!).

Therefore, n! = n × (n-1) × (n-2) × ... × 1, and is pronounced as "n factorial".

Note that by mathematical convention, 0! = 1 and 1! = 1.

The Example of Permutations

The standard track for sprinting races at the Olympics features 9 lanes. However, for the 100-meter dash, lane 1 is typically left empty. The 8 runners are instead placed in lanes 2 through 9. How many possible ways can these 8 runners be arranged across lanes 2 to 9?

Using the fundamental counting principle:

• any of the 8 runners can be assigned to lane 2,
• any of the remaining 7 runners can be assigned to lane 3,
• any of the remaining 6 runners can be assigned to lane 4,
• any of the remaining 5 runners can be assigned to lane 5,
• any of the remaining 4 runners can be assigned to lane 6,
• any of the remaining 3 runners can be assigned to lane 7,
• either of the remaining 2 runners can be assigned to lane 8,
• the 1 remaining runner is assigned to lane 9.

Therefore, the total possible permutations for arranging the 8 runners across the 8 available lanes is 8! = 8 × 7 × 6 × ... × 2 × 1 = 40,320 ways.

To solve this using our permutations calculator, simply enter 8 in both the n (objects) and r (sample) boxes, then click "Calculate" to instantly get 40,320.

Permutation of Subsets

In the previous examples, we looked at calculating permutations when every object in the set is used in the arrangement. However, there are many situations where a larger set of objects is arranged into smaller subgroups.

In these cases, the total number of available objects is denoted by n, the number of objects selected for the subgroup (the sample) is denoted by r, and the following formula is used to calculate the number of permutations:

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

This is the standard permutation formula used to calculate arrangements without repetitions when you need to organize a specific sample r taken from a larger set n.

If you need to calculate the number of ways to arrange all elements of a set in a specific order without repetitions (where n equals r), the formula simplifies to:

$$ₙPᵣ=n!$$

Example

Returning to the 100-meter dash example, we previously calculated the total number of ways all eight runners could be arranged on the track. Now, let's look at the medals. Three medals are up for grabs: first place wins gold, second place wins silver, and third place wins bronze. Out of the 8 starting runners, how many possible ways can the gold, silver, and bronze medals be awarded?

By the fundamental counting principle, any of the 8 runners can take the first-place position. Once the gold medalist is decided, 7 runners remain in contention for second place. After silver is awarded, 6 runners are left to compete for the third-place bronze medal. Therefore, the total number of possible permutations for the top three positions out of 8 runners is: 8 × 7 × 6 = 336

Alternatively, we can use the nPr formula:

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

Plugging in our numbers, we get:

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

To find this using the permutations calculator, enter 8 into the n (objects) box and 3 into the r (sample) box. Click "Calculate", and you will get 336.

Permutations and Combinations: The Difference

Another essential counting technique in mathematics is combinations. Combinations represent the various ways a smaller number of objects (sample r) can be selected from a larger pool of objects (n). The number of combinations of r objects chosen from n objects is denoted simply as ₙCᵣ.

When defining permutations, we established that the sequence or arrangement is strictly important. This is exactly the core difference between permutations and combinations: in combinations, the order does not matter.

For instance, we previously noted that the permutations of the letters XYZ taken in groups of two yield six distinct arrangements: XY, XZ, YZ, YX, ZX, and ZY.

However, the combinations of the letters XYZ in groups of two yield only three distinct pairings: XY, XZ, and YZ. Because order doesn't matter in combinations, XY and YX are considered the exact same pair. The same applies to XZ and ZX, as well as YZ and ZY.

The formula used to calculate the number of combinations of r objects from n objects is:

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

Example of Calculating Combinations

In the runner scenario above, we calculated the number of ways we could assign the specific first, second, and third-place positions from a group of 8 runners. But what if we simply want to know the number of ways to select 3 medalists from the 8 runners, regardless of their specific placement? In this case, it doesn't matter who comes in first, second, or third—only that they are selected to win a medal.

Because the exact order of the medals is irrelevant here, we use combinations. We can solve this using the standard combinations formula:

$$ₙСᵣ = \frac{n!}{r!(n-r)!}$$

The number of ways 3 unranked medalists can be selected from a pool of 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. A television news producer needs to choose 3 out of 5 available guest speakers for an upcoming analytical program. The order in which the guests speak is highly important. How many different ways can the producer arrange the presentations? Because order matters and repetition is not allowed (a guest cannot appear twice in the same lineup), we use the formula for permutations:

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

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

This shows that the producer has 60 unique ways to organize the guest speakers.

2. A prominent restaurant critic has shortlisted 10 excellent sushi establishments in town to determine the top 3 sushi restaurants. The establishments must be presented in a specific order to reflect their final ranking, and no restaurant can appear in the ranking more than once. Because the order is crucial and there are no repetitions, we satisfy the core requirements for permutations. We apply the nPr formula:

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

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

3. When we state that "order is important" in permutations, it does not strictly mean the order has to be a numerical ranking like 1st, 2nd, or 3rd. Order can also be defined by specific, distinct roles or locations to which our elements are assigned.

For example, consider the manager of a home repair company. He has four specific work orders to fulfill today: painting a visa agency office, a factory warehouse, a clothing store, and a room in a private home. The company has six painters on staff. Each painter can only be sent to 1 facility per day, meaning two painters will naturally have the day off.

The four unique work locations (the visa agency, warehouse, store, and private home) serve as the equivalents of positions 1, 2, 3, and 4.

The manager evaluates his options:

• 6 available painters who can be assigned to the visa agency office,
• 5 remaining painters to be assigned to the factory warehouse,
• 4 remaining painters to be sent to the clothing store,
• 3 remaining painters who can be assigned to the private home.

Intuitively, we can calculate the total number of assignment choices as 6 × 5 × 4 × 3 = 360.

Because the specific location each painter is assigned to matters entirely (order is important), and no painter can work at more than one location in a single day (no repetitions), we can perfectly apply our permutation formula:

$$ₙPᵣ = \frac{n!}{(n-r)!}$$

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

Ultimately, there are exactly 360 different ways the home repair manager can allocate the day's orders among his available painters under the given conditions.