In this article, we will examine the world of algorithms. Specifically, we will talk about an algorithm that we can use to generate permutations.

To download the source code for this article, you can visit our GitHub repository.

Let’s start.

## Permutations

Firstly, we need a definition for permutation, of which there are a few. Even Wikipedia’s definition is quite difficult to understand. Therefore, let us **say that a permutation is an arrangement of items in a particular order, with the order being paramount**.

To dive in a little further, let’s examine a concrete example. Let us say that we have three coins (`1`

, `2`

, and `5`

cents). What are the different possibilities for arranging these coins? It turns out that there are six possible arrangements: `[1, 2, 5]`

, `[1, 5, 2]`

, `[2, 1, 5]`

, `[2, 5, 1]`

, `[5, 1, 2]`

, `[5, 2, 1]`

. This is the set of all possible **permutations** of 3 items.

### Number of Permutations

Before moving on to programming, let us answer a theoretical question: “How many possible permutations are there for N items?’. We already saw that there are six possibilities when we have three items.

We have N options for the first position if we have N items. Then we only have N-1 items for the second position, as we already used 1 for the first position. On the third position, we have N-2 items, and so on. With that, we can deduce a formula as N * (N-1) * (N-2) * … * 3 * 2 * 1, which is a mathematical definition for **factorial,** and the symbol for factorial is an exclamation mark (!).

**The number of all possible permutations on N items is N!**, which can quickly become very large. For example, with three items, there are six permutations. Moreover, if we have six items, there are already 720 permutations. Similarly, with ten items, there are 3,628,800 permutations; with 20 items, there are whooping 2,432,902,008,176,640,000 permutations.

To put this into perspective, suppose we have a printer so fast that it can print 1000 permutations in one second. That printer would print all permutations of the 20 elements for about 77 million years. Keeping that in mind, we’ll want to use an algorithm as fast as possible.

### How Excellent and Quick Is an Algorithm? The Big O Notation.

In the field of combinatorics and algorithms, determining an algorithm’s quality is crucial. We need to classify whether algorithm A is better than algorithm B. **The “Big O” notation is a special notation in computer science that helps determine the upper bound of the number of calculations an algorithm must execute to achieve a result.**

The “Big O” notation classifies algorithms into classes such as **n, n^2, log(n), n*log(n), etc.** These classifications indicate that their algorithms will have to execute, for example, log(n) calculations to produce a result for n elements. The binary search algorithm is an excellent example of such a log(n) algorithm.

In our case, however, there is not much to calculate. We have already established that the number of permutations for n elements is n!. Even the best algorithm in the universe can’t solve it without executing at least n! calculations, so we can’t do any better. But we can do a bit worse. First, we’ll write the most straightforward algorithm possible, with much higher complexity (n^n) and lower performance.

## Basic Permutations Algorithm

The basic algorithm is simple and can be visualized as a bicycle lock with rolling numbers. To find out all the permutations of three digits, we can choose all possible positions of all three wheels on the lock, and we will get this: `000`

, `001`

, `002`

, `010`

, `011`

, `012`

, `020`

, `021`

, `022`

, `100`

, `101`

, `102`

, `110`

, `111`

, `112`

, `120`

, `121`

, `122`

, `200`

, `201`

, `202`

, `210`

, `211`

, `212`

, `220`

, `221`

, and `222`

.

In other words, there are 3^3=27 positions or three loops with three elements:

public static void PrintAllPossiblePositions(){ int counter = 1; for (int p0 = 0; p0 < 3; p0++) for (int p1 = 0; p1 < 3; p1++) for (int p2 = 0; p2 < 3; p2++) Console.WriteLine($"Position [{counter++, 2}]: {p0}{p1}{p2}");}

But not all results are valid. With permutations, there can’t be repeating digits. So from this set of 27 results, we must discard every result with repeating digits. By applying that, we will get the correct result: `012`

, `021`

, `102`

, `120`

, `201`

, and `210`

.

To print out only permutations, we have to discard results with repeating digits:

public static void PrintPermutations(){ int counter = 1; for (int p0 = 0; p0 < 3; p0++) for (int p1 = 0; p1 < 3; p1++) for (int p2 = 0; p2 < 3; p2++) { if ((p0 != p1) && (p0 != p2) && (p1 != p2)) Console.WriteLine($"Permutation [{counter++,2}]: {p0}{p1}{p2}"); }}

This method will print the permutations of 3 elements.

So, we have written our first working algorithm for generating permutations. **However, as discussed in the beginning, the algorithm is inefficient since we calculate 27 elements to produce just six results.**

### Algorithm Improvement

We can improve a lot with a small algorithm change.

Let’s take these arrangements: `000`

, `001`

, `002`

. We can discard all three arrangements in the second loop. **As soon as we see another 0, we can quit**, as there can’t be two zeros in a given permutation:

public static void PrintPermutationsImproved(){ int counter = 1; for (int p0 = 0; p0 < 3; p0++) for (int p1 = 0; p1 < 3; p1++) { if (p1 == p0) continue; // Just skip the whole inner loop(!) for (int p2 = 0; p2 < 3; p2++) { if ((p2 == p0) || (p2 == p1)) continue; Console.WriteLine($"Permutation [{counter++,2}]: {p0}{p1}{p2}"); } }}

We avoid executing the third loop when we discover that the second position would have the same digit as the first. This algorithm is much better than the first one, **but they both have one major drawback! **

They are useless for generating permutations of 4 or 6 elements. To display all the permutations of 4 or 6 elements, we need a method with 4 or 6 for loops. This will require a different method for each N, making this approach less than ideal.

## Recursion to Save the Day

**Fortunately, many problems, especially combinatorial problems, can be solved using recursion.** We have to find how a smaller N fits a bigger N. So to use recursion, we will need to find out how to generate permutations of 4 elements (or N elements) if we know how to build permutations with three elements (or N-1 elements).

We already have permutations of 3 elements: `012`

, `021`

, `102`

, `120`

, `201`

, and `210`

. If we add element 3, what do we get? Well, for starters, we should get `3012`

, `3021`

, `3102`

, `3120`

, `3201`

, `3210`

, and then more of them. So we have created new permutations by adding a new element to existing permutations, which is key to the solution!

First, we have a bag of elements. Then, we take the first element out of the bag and create all the permutations of the remaining elements in the bag. We paste the first element to every generated permutation. Next, we return the first element to the bag and take the next element out. We repeat this process until we have used all the elements in the bag.

To illustrate this idea, let’s try generating permutations for four elements using the algorithm.

### The Recursive Algorithm

Let’s say we have a bag of four elements: `0`

, `1`

, `2`

, and `3`

. We take the first element out of the bag, which is `0`

, and create permutations of the remaining elements (`1`

, `2`

, and `3`

). We generate six permutations: `123`

, `132`

, `213`

, `231`

, `312`

, and `321`

. To obtain the final permutations, we add `0`

to the beginning of each generated permutation: `0123`

, `0132`

, `0213`

, `0231`

, `0312`

, and `0321`

. We then return `0`

to the bag and take the next element out, which is `1`

. The bag now contains elements `0`

, `2`

, and `3`

; their permutations are `023`

, `032`

, `203`

, `230`

, `302`

, and `320`

. Now add `1`

to the beginning of each permutation to obtain: `1023`

, `1032`

, `1203`

, `1230`

, `1302`

, and `1320`

. Repeat the same process for elements `2`

and `3`

.

Ultimately, we have used all four elements in the bag and generated all possible permutations. The total number of permutations generated is 24 (4 * 6), which is the correct result.

The code written in C# follows the algorithm:

public static void RecursiveAlgorithm(string elementsOutOfTheBag, List<int> elementsInTheBag){ if (elementsInTheBag.Count == 0) Console.WriteLine($"Permutation: {elementsOutOfTheBag}"); else { foreach (var element in elementsInTheBag) { List<int> newBag = elementsInTheBag.Where(e => e != element).ToList(); RecursiveAlgorithm(elementsOutOfTheBag + element, newBag); } }}

The method accepts the string of already positioned elements and a bag of not-yet-used elements. If the bag is empty, we have reached the end and have the resulting permutation. Otherwise, we recursively execute the same algorithm with the next element.

### Where to Begin

To initiate this recursive method, we require a starting position. In this case, it is an empty string and all the elements:

public static void PrintPermutationsRecursively(){ RecursiveAlgorithm("", new List<int> { 0, 1, 2, 3, 4 });}

The result will be 120 permutations for five elements.

**However, we should note that this method does not have a limit on the number of elements**. This simple and easy-to-understand method produces a working result for any quantity desired. We can generate permutations for eight elements by simply calling:

`RecursiveAlgorithm("", new List<int> { 0, 1, 2, 3, 4, 6, 7, 8 });`

Executing this code generates 40,320 permutations, keeping our algorithm busy for a while.

### Breaking the Recursion

Although recursion is an essential technique, there is always a potential problem with it. **If a long recursion isn’t allowed to exit, it can become a severe problem**. Therefore, as programmers, we must be sure our algorithm provides a way out.

In our case, this is relatively easy to check because, with every recursion, we take one more element out of the bag. Eventually, we will remove all the elements, and the recursion will end.

It’s always a good idea to write tests to verify whether an algorithm provides the correct result, and you can find our tests in the GitHub project.

### Recursive Performance

**To determine our algorithm’s performance, we will have to measure it.** We will use a standard tool: BenchmarkDotNet.

To benchmark different algorithms fairly, we have to strip everything nonessential from them. In our case, we will rewrite the algorithms so that they will not write results to the console, and they will permutate elements in the same memory location.

The second requirement regarding memory location may not be immediately apparent, but it is essential. Since there are many permutations, it makes sense to generate all permutations in the same memory location and not utilize memory unnecessarily for each new permutation.

To this end, we will also introduce a new parameter: *benchmark*. If this parameter is true, we will elect to “forget” each permutation. We will generate it and move to the next one. This way, measuring speed and memory consumption will be fairer.

### Rewriting the Recursive Algorithm

By stripping the console output from the algorithm and collecting the results to list, we will get our next iteration of the recursive algorithm:

private void RecursiveGenerator(List<byte[]> result, byte[] input, byte fixedPosition, List<byte> freeNumbers, bool banchmarks){ if (fixedPosition == input.Length) { if (!banchmarks) result.AddCopy(input); } else { for(int i = 0; i < freeNumbers.Count; ++i) { var newFreeNumbers = new List<byte>(freeNumbers); newFreeNumbers.RemoveAt(i); input[fixedPosition] = freeNumbers[i]; RecursiveGenerator(result, input, (byte)(fixedPosition + 1), newFreeNumbers); } }}

The algorithm is exactly the same as before; only the parameters are slightly different.

The first parameter is a list that will return the resulting permutations. The second parameter is a memory location where the algorithm will generate permutations. So every permutation will occupy the same memory (the same array), and before we add it to the resulting list, we have to make a copy! The third parameter counts how many elements in the array we have already constructed, and the algorithm will stop when we generate the whole permutation. The fourth parameter is a bag of unused elements, just as before. The last parameter is a flag that tells the algorithm if we are measuring time/space, so in that case, it will not generate the resulting output.

### Rewriting the Basic and Improved Basic Algorithms

In addition, we also need to rewrite the Basic and Improved Algorithms. Specifically, we will focus on the methods used to generate permutations of 3 elements. Given that the code for generating five elements using the Basic or Improved algorithm is quite lengthy, here we will only modify the code for generating permutations of 3 elements:

private List<byte[]> GetPermutations3(byte[] input, bool doBanchmarks){ var result = new List<byte[]>(); var number = input.Length; for (byte i0 = 0; i0 < number; ++i0) for (byte i1 = 0; i1 < number; ++i1) for (byte i2 = 0; i2 < number; ++i2) if ((i0 != i1) && (i0 != i2) && (i1 != i2)) { input[0] = i0; input[1] = i1; input[2] = i2; if (!doBanchmarks) result.AddCopy(input); } return result;}

To do so, we create an array to store the results and include a flag to indicate if we are measuring the code’s performance. The algorithm remains largely the same, with the only changes made to improve efficiency using the same memory location.

We can apply these same changes to the Improved algorithm as well:

private List<byte[]> GetPermutations3(byte[] input, bool doBanchmarks){ var result = new List<byte[]>(); var number = input.Length; for (byte i0 = 0; i0 < number; ++i0) { input[0] = i0; for (byte i1 = 0; i1 < number; ++i1) { if (i1 == i0) continue; input[1] = i1; for (byte i2 = 0; i2 < number; ++i2) { if ((i2 == i0) || (i2 == i1)) continue; input[2] = i2; if (!doBanchmarks) result.AddCopy(input); } } } return result;}

## Comparing Algorithms

**By making these changes, we can finally measure the algorithms’ performance.** Specifically, we will test their ability to generate permutations of 2, 4, 6, and 8 elements. The results of our testing on a typical modern desktop computer are:

Method | Items | Mean | Allocated |
---|---|---|---|

BasicAlgorithm | 2 | 69.11 ns | 176 B |

BetterBasicAlgorithm | 2 | 68.05 ns | 176 B |

RecursiveAlgorithm | 2 | 238.47 ns | 496 B |

BasicAlgorithm | 4 | 437.24 ns | 176 B |

BetterBasicAlgorithm | 4 | 254.87 ns | 176 B |

RecursiveAlgorithm | 4 | 2,605.47 ns | 4336 B |

BasicAlgorithm | 6 | 150,842.08 ns | 176 B |

BetterBasicAlgorithm | 6 | 25,801.66 ns | 176 B |

RecursiveAlgorithm | 6 | 77,412.62 ns | 125424 B |

BasicAlgorithm | 8 | 34,504,260.23 ns | 214 B |

BetterBasicAlgorithm | 8 | 3,191,882.23 ns | 178 B |

RecursiveAlgorithm | 8 | 4,196,019.67 ns | 7014645 B |

From this table, it is evident that the Improved algorithm is the fastest. Additionally, we can see that as the number of elements to be permuted increases, the Basic algorithm’s efficiency decreases while the recursive algorithm’s efficiency improves. **However, we must also consider that the recursive algorithm significantly impacts memory.**

Therefore, the recursive algorithm appears not as optimal as we initially thought. While it is currently the only algorithm capable of generating permutations for any number of elements, it is not ideal. **On the other hand, if we want an Improved algorithm for 12 elements, we would need to create a method with 12 for loops, which is impractical.**

### Raising the Bar

Can we do better? Certainly! While the three previous algorithms were relatively easy to write, they were not the best possible solutions. More advanced and efficient algorithms have been developed, such as Heap’s Algorithm for generating permutations.

## Generate Permutations With the Heap’s Algorithm

This algorithm, named after B. R. Heap, who published it in 1963, is also recursive. It involves swapping elements in a set and recursively generating permutations of the smaller set until all permutations have been generated. Fortunately, we have all the tools we need to implement Heap’s Algorithm in C#:

private void Heaps(List<byte[]> result, byte[] input, byte size, bool banchmarks){ if (size == 1) { if (!banchmarks) result.AddCopy(input); } else { for (int i = 0; i < size; i++) { Heaps(result, input, (byte)(size - 1)); if (size % 2 == 1) (input[size - 1], input[0]) = (input[0], input[size - 1]); else (input[size - 1], input[i]) = (input[i], input[size - 1]); } }}

**Despite being short and having the same parameters as our previous recursive algorithm, Heap’s Algorithm is significantly faster and more memory-efficient**, as it avoids the need to generate a bag during every recursion.

The theory behind Heap’s algorithm is described in great detail on Wikipedia, so we will not repeat it here.

## Conclusion

The fact that we are unlikely to find better algorithms than some of the world’s great programming minds have already discovered does not mean we should give up. We should consider the various possibilities, write different algorithms and ideas, and not always settle for “the best algorithm” available online. In truth, there is rarely a “best one” for every case, as it depends on the specific problem we have to solve and the type of data it entails.

In this case, we have demonstrated how to approach the problem of generating permutations, and once we have our starting point, we can continue to seek out and craft ever-better algorithms.

## FAQs

### How to calculate permutations in C#? ›

The total number of permutations for order n is factorial(n), usually written as n! and calculated as **n * (n-1) * (n-2) * . . 1**. For example, 5! = 5 * 4 * 3 * 2 * 1 = 120. The number of permutations gets very, very large as n increases.

**How to find all permutations of an array in C#? ›**

- We start by iterating through each element of our input array.
- Swap the elements within nums at i and start (the first element is fixed on the first iterations), and then recursively call DoPermute on the remaining elements.
- Swap the previously swapped elements back for the next iteration (backtrack).

**What is the fastest algorithm to generate permutations? ›**

**Heap's algorithm** generates all possible permutations of n objects. It was first proposed by B. R. Heap in 1963. The algorithm minimizes movement: it generates each permutation from the previous one by interchanging a single pair of elements; the other n−2 elements are not disturbed.

**What is the formula for permutation in coding? ›**

**P _{k} = n! / (n - k)!**

**How to count occurrences of number in array in C#? ›**

You can count the total number of elements or some specific elements in the array **using an extension method Count() method**. The Count() method is an extension method of IEnumerable included in System. Linq. Enumerable class.

**How do you generate permutations of an array? ›**

**Algorithm**

- Make a vector nums that will store the current permutation.
- Make a set of vectors and push the vector nums into it.
- At last, print the content of this set; this will give us all possible permutations of the given array without any duplicates.

**How do you print all permutations of an array? ›**

Example 1: Input: arr = [1, 2, 3] Output: [ [1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1] ] Explanation: **Given an array, return all the possible permutations**. Example 2: Input: arr = [0, 1] Output: [ [0, 1], [1, 0] ] Explanation: Given an array, return all the possible permutations.

**How do you generate a random permutation efficiently? ›**

A simple algorithm to generate a permutation of n items uniformly at random without retries, known as the Fisher–Yates shuffle, is to start with any permutation (for example, the identity permutation), and then go through the positions 0 through n − 2 (we use a convention where the first element has index 0, and the ...

**How do you find all possible combinations? ›**

Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula **nCr = n! / r!** *** (n - r)!**, where n represents the total number of items, and r represents the number of items being chosen at a time.

**Which command generate random permutations? ›**

**p = randperm( n )** returns a row vector containing a random permutation of the integers from 1 to n without repeating elements.

### What program do you use to find all permutations of a number? ›

**Algorithm**

- STEP 1: START.
- STEP 2: DEFINE n, r, per, fact1, fact2.
- STEP 3: PRINT n, r.
- STEP 4: fact1 =n.
- STEP 5: REPEAT STEP 6 UNTIL i>=1.
- STEP 6: fact1 = fact1*i.
- STEP 7: DEFINE number.
- STEP 8: SET number = n - r.

**What is the complexity to generate all permutations? ›**

The time complexity of this approach is **O(N!)**, Where N is the length of the given string. Reason: The recursive function uses the O(N) recursion stack. We also store the permutations in a list that occupies O(N!)

**What are the permutations of 1 2 3 4? ›**

There are **24 permutations**. The 12 even permutations are: id , (1 2 3 4) , (1 3 2 4) , (1 4 2 3) , (1 2 3) , (1 2 4) , (1 3 2) , (1 3 4) , (1 4 2) , (1 4 3) , (2 3 4) , (2 4 3). The 12 odd ones: (1 2) , (1 3) , (1 4) , (2 3) , (2 4) , (3 4) , (1 2 3 4) , (1 2 4 3) , (1 3 2 4) , (1 3 4 2) , (1 4 2 3) , (1 4 3 2).

**How do you find permutations without repetition? ›**

The number of permutations without repetitions is: ** ^{n}P_{r} = (n!) / (n - r)!**. The number of permutations with repetitions is: n

^{r}. The number of permutations around a circle is (n - 1)!. The number of permutations if there are 'r' same things, 's' same things, and 'p' same things out of 'n' total things is: n! / (r!

**What are the ways to calculate permutation? ›**

There is a way you can calculate permutations using a formula. This formula is: **P(n,r) = n!** **÷ (n-r)!**

**Is permutation and combination hard? ›**

Every topic in Mathematics is easy if you practice and understand its logic. Permutation and combination is a topic that requires logical thinking. **Permutation and combination topic is easier as compared to other topics in mathematics such as calculus**.

**What is the difference between array count and length in C#? ›**

**A collection's Count is the number of items currently stored in it and can potentially change over time.** **An array's Length is the maximum number of items it can hold** (it will have a length of 10 even if you haven't stored that many items in it) and is immutable.

**How do arrays work in C#? ›**

In C#, an array is **a structure representing a fixed length ordered collection of values or objects with the same type**. Arrays make it easier to organize and operate on large amounts of data. For example, rather than creating 100 integer variables, you can just create one array that stores all those integers!

**How many permutations does 4 numbers have? ›**

In our example, we want to know the number of permutations of 4 digits. By our rule, this number is equal to **4!**. We get that there are 24 possible permutations of 4 digits, when repetition is not allowed.

**How do you find all permutations of an array using recursion? ›**

Take the first element of the array (1), and set it aside. Take the remaining elements (2, and 3) and make one permutation with the original order (1, 2, 3), and another one with the original remaining elements (2 , 3), but swapped (3, 2). So now we have two permutations: [1, 2, 3] and [1, 3, 2].

### How many permutations will be formed from the array arr ={ 1 2 3 4? ›

3. What will be the lexicographical order of permutations formed from the array arr={1,2,3}? Explanation: The number of permutations for the problem will be **6** according to the formula 3P3.

**How do you print the most repeated number in an array? ›**

**Algorithm**

- Step 1 − Declare and initialize an integer array.
- Step 2 − Take an int variable max_count and initialize it with 0. ...
- Step 3 − Check for condition count > max_count. ...
- Step 4 − Finally, print the element which has max_count value.

**How to print everything in an array C? ›**

Step 1 → Take the array arr[] and define its values, that is define the value for each element of the array arr[]. Step 2 → Start a loop for each value of the array arr[]. Step 3 → Print the value associated with arr[i] where i is the current iteration.

**What is recursion in permutations? ›**

Permutations, which are arrangements of a set of elements in a specific order. **Using recursion, we can generate all possible permutations of a given set, allowing us to examine every possible combination of elements in a specific order**.

**Which algorithm is best for permutation? ›**

**Heap's Algorithm**

One of the more traditional and effective algorithms used to generate permutations is the method developed by B. R. Heap. This algorithm is based on swapping elements to generate the permutations. It produces every possible permutation of these elements exactly once.

**What algorithm generates randomness? ›**

**A pseudorandom number generator (PRNG), also known as a deterministic random bit generator (DRBG)**, is an algorithm for generating a sequence of numbers whose properties approximate the properties of sequences of random numbers.

**What are the possible combinations of 1 2 3 4 5 6? ›**

There are **720** permutations of the digits 1,2,3,4,5,6 supoose these permutations are arranged from smallest to largest numerical values beginning from 123456 and ending with 654321.

**How to generate random number in C# console application? ›**

**The Random class provides the following methods to generate random numbers.**

- The Next() method. This method generates a random positive number within the range -2,147,483,648 to 2,147,483,647. ...
- The Next(int) method. ...
- The Next(int, int) method. ...
- The NextDouble() method.

**How to generate a random permutation with random transposition? ›**

The usual algorithm for generating a random permutation goes as follows: **Choose a random integer U 1 uniformly between 1 and n, then transpose U 1 and 1.** **At stage j, choose a random integer Uj uniformly between j and n, then transpose Uj and j**.

**How many permutations does an 8 digit number have? ›**

We cannot put 0 in the first place digit because then it will become a 7 digit number. Now, we know that for the rest of the 7 digits in the 8 digit number there are **10 choices available that are from 0 to 9**. Hence, there are a total of 90,000,000 8-digit numbers.

### How to get all permutations in C? ›

**C program to find permutations of given strings**

- if s[i] > s[i - 1]), then: j := i + 1.
- for j < n, update (increase j by 1), do: if s[j] <= s[i - 1]), then: Come out from the loop. ...
- s[i - 1] = s[j - 1]
- s[j - 1] = t.
- for i < n - 1, update (increase i by 1), (decrease n by 1), do: t := s[i] s[i] := s[n - 1] ...
- return 1.

**Why are there 999 permutations? ›**

The reason we use 999, or 9999, or 99999, permutations in that random subsample is that **the observation you have is a possible permutation, no matter how unlikely it may be if the null hypothesis is true**. This then gives you a nice round probability (p value) of 1/1000, 1/10000, 1/100000 etc.

**How many permutations are there in success? ›**

The rank of the word SUCCESS, If all possible permutations of the word SUCCESS are arranged in dictionary order is **331**. The alphabetic letters in Success are S, U, C, and E.

**How do you manually solve permutations? ›**

To calculate the number of permutations, **take the number of possibilities for each event and then multiply that number by itself X times, where X equals the number of events in the sequence**. For example, with four-digit PINs, each digit can range from 0 to 9, giving us 10 possibilities for each digit.

**Which has more outcomes permutation or combination? ›**

In the calculation of “how many (permutations),(combinations) can be made from K objects out of N candidates, there will be **more permutations than combinations**, because each combination can be rearranged to make many permutations.

**How to count the number of each letter in a string C#? ›**

We can **use System.** **Linq to count the number of characters using the Count() method**. The Count() method of Linq iterates over every character of the string and executes the predicate, in our case t => t == toFind for each of them.

**How to count the number of entries in C#? ›**

You can count the total number of elements or some specific elements in the array **using an extension method Count() method**. The Count() method is an extension method of IEnumerable included in System.

**What is a permutation in CS? ›**

A permutation is **a one-to-one function from a set onto itself**. Permutations. A permutation is a function that rearranges the order of terms in a sequence. It is useful to study a few small examples. In computing practice, sorting a group of objects into a preferred order is a fundamental operation.

**How do you find the number of permutations in a string? ›**

The number of permutations of a string is **always equal to the factorial of the length of the string**. For example : string HELLO of length 5 has 5! permutations i.e. 120 permutations in total.

**How to find multiple occurrences of a string in C#? ›**

**Count occurrences of a character within a string in C#**

- Using Enumerable.Count() method ( System.Linq ) The recommended solution is to use LINQ's Count() method to count occurrences of the given character in a string. ...
- Using Enumerable.Where() method ( System.Linq ) ...
- Using String.Split() method.

### How to count repeated words in a string C#? ›

**Repeated Word Count In A String Using C# Dictionary**

- using System;
- using System.Collections.Generic;
- using System.Linq;
- using System.Text;
- using System.Threading.Tasks;
- namespace CountRepeatedWordCount.
- {
- class Program.

**How do you find the number of repeating characters in a string? ›**

To find the duplicate character from the string, we **count the occurrence of each character in the string**. If count is greater than 1, it implies that a character has a duplicate entry in the string. In above example, the characters highlighted in green are duplicate characters.

**What are the 4 types of permutations? ›**

The four types of permutations are **permutations with repetition, permutations without repetition, permutations with multi-sets, and circular permutations**.

**What is 1 example of permutation? ›**

For example, if you have a lock where you need to enter four digits, the order matters. If the correct numbers are 8 3 6 2, you can't enter the same numbers in any other order (e.g., 6 8 2 3) and expect the lock to open! Hence, that's a permutation.

**How many permutations does a 5 digit code have? ›**

Assuming no five-digit number can begin with zero, there are 9 possible choices for the first digit. Then there are 10 possible choices for each of the remaining four digits. Therefore, you have 9 x 10 x 10 x 10 x 10 combinations, or 9 x 10^4, which is **90,000 different combinations**.

**How many permutations does a 6 digit pin have? ›**

Answer and Explanation: The combinations for this problem depends on whether repetition of digits is allowed or not. If repetition is not allowed, then there are 10 * 9 * 8 * 7 * 6 * 5 possible combinations = **151200**.