Permutation (shortened as p(n,r)) and combination (abbreviated as c(n,r)) are methods for representing a collection of things by picking them from a set and dividing them into subsets. It specifies the numerous methods for organizing a set of data. Permutations are used to pick data or objects from a group, whereas combinations are used to represent the manner in which they would be displayed. In mathematics, both concepts are crucial.

Permutation and combination, as well as the differences between them, are thoroughly detailed here. In this article, you will get a brief understanding of p(n,r) and c(n,r), their definitions, the way to calculate the p(n,r) status, i.e., the p(n,r) value, as well as the c(n,r) value.

## Permutation Meaning

A permutation is a set of objects arranged in such a way that the order matters. Consider an example of the number of permutations of n objects taken n at a time. When counting without replacement and considering that the order is essential, we employ permutation and determine the p(n,r) status. We can utilize combinations if the order doesn’t matter.

P(n, r) represents the range of permutations of n items taken r at a time in general. These sorts of permutation problems can be solved using either the permutation formula or reasoning.

The formulas for p(n,r), c(n,r), and permutation with repeated symbols are shown below. More examples and step-by-step solutions of finding p(n,r) status can be found further down the page.

- Permutations of n things, taking r at a time when order matters: p(n,r) = (n!) / ((n-r)!)
- Combinations of n things, taking r at a time when order does not matter: c(n,r) = (n!) / ((n-r)! (r!))
- Permutations of n things, with n1 items repeated, n2 items repeated, and so on up to nk items: p(n,r) = (n!) / ((n1)! (n2)! …(nk!))

### Deriving p(n,r) status formula

Because a permutation entails selecting r separate things without replacing any of the n items, and because the order is necessary, using the fundamental counting principle, we get P (n, r) = n. (n-1). (n-2). (n-3) ……

There are (n-(r-1)) methods to do it.

It can be expressed in the following way:

P (n, r) = n .(n-1) . (n-2). (n-3)…. (n-r+1) ———————-> (1)

We get 3. 2. 1 by multiplying and dividing the above equation (1) by (n-r) (n-r-1) (n-r-2).

P (n, r) = {n .(n-1) . (n-2) ….. (n-r+1) [(n-r)(n-r-1)(n-r-2) ….. 3.2.1] } / [(n-r)(n-r-1)(n-r-2)…..3.2.1]

P (n, r) = n! / (n-r)!

### Deriving C(n,r) Formula

Because combinations include selecting r things from a set of n objects with no regard for their order, we can deduce that:

C(n,r) denotes the number of permutations/possibilities for arranging r objects. [We know that the range of possibilities to arrange r things in r ways = r because of the fundamental counting concept!]

C (n,r) = P (n, r) / r!

C (n,r) = n! / [(n−r)! r!]

As a result, C (n,r) =n! / [(n−r)! r!]

### Examples of P (n,r) and C (n,r)

Assume there are 9 girls and 14 boys in the class. If the teacher needs to choose a girl or a boy to be the class monitor, he or she might choose among 14 boys or 9 girls. She can accomplish it in a total of 14 + 9 = 23 different ways (using the sum rule of counting). She can achieve it in a total of 14 + 9 = 23 different ways (using the sum rule of counting).

Let’s take a look at a different case. Assume Sam orders one main meal and a drink regularly. Today, he can have a pizza, burger, watermelon juice, hot dog, or orange juice. What are all the different combinations he may try? There are three snack options and two beverage options. To find the combinations, we multiply. 6 is the result of 3 x 2 = 6. Using the product rule of counting, Sam can attempt 6 different combinations.

## Permutations vs. Combinations

Permutation (p(n,r) status) | Combination (c(n,r) status) |

For the arrangement of digits, people, alphabets, numbers, colors, letters, etc. | For selecting a food, menu, subjects, clothes, team, etc. |

Choosing a pitcher, team captain, and shortstop from a pool of candidates. | Choosing three people from a group to form a team. |

Choosing the two most popular colors from a color pamphlet in sequence. | Choosing two colors out of a color catalog. |

Choosing the winners of the first, middle, and third places. | Choosing three winners. |

### Conclusion

When calculating the range of possible combinations in a collection, it might be hard to tell the difference between permutations and combinations. Generally, to estimate the range of possible combinations of a group of items, the process of permutations is used. In contrast, we can ‘select’ from a wider set of objects when we use combinations to count the number of combinations. Permutations and combinations are crucial statistical concepts.