Permute 3 123
- The top 3 will receive points for their team. How many different permutations are there for the top 3 from the 12 contestants? For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3.
- A common task in programming interviews (not from my experience of interviews though) is to take a string or an integer and list every possible permutation.
Calculator Use
Like the Combinations Calculator the Permutations Calculator finds the number of subsets that can be taken from a larger set. However, the order of the subset matters. The Permutations Calculator finds the number of subsets that can be created including subsets of the same items in different orders.
- Factorial
- There are n! ways of arranging n distinct objects into an ordered sequence, permutations where n = r.
- Combination
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are not allowed.
- Permutation
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are not allowed. When n = r this reduces to n!, a simple factorial of n.
- Combination Replacement
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does not matter and replacements are allowed.
- Permutation Replacement
- The number of ways to choose a sample of r elements from a set of n distinct objects where order does matter and replacements are allowed.
- n
- the set or population
- r
- subset of n or sample set
Permutations Formula:
The top 3 will receive points for their team. How many different permutations are there for the top 3 from the 12 contestants? For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3. Mac软件：Permute 3 v3.5.16 最纯粹简洁的多媒体转码软件 2021年02月09日 5721点热度 5人点赞 2条评论 Permute是最容易使用的媒体转换器，它易于使用，无需配置，拖放界面，它将满足转换所有媒体文件的需求。.
For n ≥ r ≥ 0.
Calculate the permutations for P(n,r) = n! / (n - r)!. 'The number of ways of obtaining an ordered subset of r elements from a set of n elements.'[1]
Permutation Problem 1
Choose 3 horses from group of 4 horses
In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). So out of that set of 4 horses you want to pick the subset of 3 winners and the order in which they finish. How many different permutations are there for the top 3 from the 4 best horses?
For this problem we are looking for an ordered subset of 3 horses (r) from the set of 4 best horses (n). We are ignoring the other 11 horses in this race of 15 because they do not apply to our problem. We must calculate P(4,3) in order to find the total number of possible outcomes for the top 3 winners.
P(4,3) = 4! / (4 - 3)! = 24 Possible Race Results
If our 4 top horses have the numbers 1, 2, 3 and 4 our 24 potential permutations for the winning 3 are {1,2,3}, {1,3,2}, {1,2,4}, {1,4,2}, {1,3,4}, {1,4,3}, {2,1,3}, {2,3,1}, {2,1,4}, {2,4,1}, {2,3,4}, {2,4,3}, {3,1,2}, {3,2,1}, {3,1,4}, {3,4,1}, {3,2,4}, {3,4,2}, {4,1,2}, {4,2,1}, {4,1,3}, {4,3,1}, {4,2,3}, {4,3,2}
Permutation Problem 2
Choose 3 contestants from group of 12 contestants
At a high school track meet the 400 meter race has 12 contestants. The top 3 will receive points for their team. How many different permutations are there for the top 3 from the 12 contestants?
For this problem we are looking for an ordered subset 3 contestants (r) from the 12 contestants (n). We must calculate P(12,3) in order to find the total number of possible outcomes for the top 3.
P(12,3) = 12! / (12-3)! = 1,320 Possible Outcomes
Permutation Problem 3
Choose 5 players from a set of 10 players
An NFL team has the 6th pick in the draft, meaning there are 5 other teams drafting before them. If the team believes that there are only 10 players that have a chance of being chosen in the top 5, how many different orders could the top 5 be chosen?
For this problem we are finding an ordered subset of 5 players (r) from the set of 10 players (n).
P(10,5)=10!/(10-5)!= 30,240 Possible Orders
References
[1] For more information on permutations and combinations please see Wolfram MathWorld: Permutation.
Introduction
Suppose we have a finite sequence of numbers like (0, 3, 3, 5, 8), and want to generate all its permutations. What is the best way to do so?
The naive way would be to take a top-down, recursive approach. We could pick the first element, then recurse and pick the second element from the remaining ones, and so on. But this method is tricky because it involves recursion, stack storage, and skipping over duplicate values. Moreover, if we insist on manipulating the sequence in place (without producing temporary arrays), then it’s difficult to generate the permutations in lexicographical order.
It turns out that the best approach to generating all the permutations is to start at the lowest permutation, and repeatedly compute the next permutation in place. The simple and fast algorithm for performing this is what will be described on this page. We will use concrete examples to illustrate the reasoning behind each step of the algorithm.
The algorithm
We will use the sequence (0, 1, 2, 5, 3, 3, 0) as a running example.
The key observation in this algorithm is that when we want to compute the next permutation, we must “increase” the sequence as little as possible. Just like when we count up using numbers, we try to modify the rightmost elements and leave the left side unchanged. For example, there is no need to change the first element from 0 to 1, because by changing the prefix from (0, 1) to (0, 2) we get an even closer next permutation. In fact, there is no need to change the second element either, which brings us to the next point.
Firstly, identify the longest suffix that is non-increasing (i.e. weakly decreasing). In our example, the suffix with this property is (5, 3, 3, 0). This suffix is already the highest permutation, so we can’t make a next permutation just by modifying it – we need to modify some element(s) to the left of it. (Note that we can identify this suffix in Θ(n) time by scanning the sequence from right to left. Also note that such a suffix has at least one element, because a single element substring is trivially non-increasing.)
Secondly, look at the element immediately to the left of the suffix (in the example it’s 2) and call it the pivot. (If there is no such element – i.e. the entire sequence is non-increasing – then this is already the last permutation.) The pivot is necessarily less than the head of the suffix (in the example it’s 5). So some element in the suffix is greater than the pivot. If we swap the pivot with the smallest element in the suffix that is greater than the pivot, then the prefix is minimized. (The prefix is everything in the sequence except the suffix.) In the example, we end up with the new prefix (0, 1, 3) and new suffix (5, 3, 2, 0). (Note that if the suffix has multiple copies of the new pivot, we should take the rightmost copy – this plays into the next step.)
Finally, we sort the suffix in non-decreasing (i.e. weakly increasing) order because we increased the prefix, so we want to make the new suffix as low as possible. In fact, we can avoid sorting and simply reverse the suffix, because the replaced element respects the weakly decreasing order. Thus we obtain the sequence (0, 1, 3, 0, 2, 3, 5), which is the next permutation that we wanted to compute.
Condensed mathematical description:
Find largest index i such that array[i − 1] < array[i].
(If no such i exists, then this is already the last permutation.)Find largest index j such that j ≥ i and array[j] >array[i − 1].
Swap array[j] and array[i − 1].
Reverse the suffix starting at array[i].
Overall, this algorithm to compute the next lexicographical permutation has Θ(n) worst-case time complexity, and Θ(1) space complexity. Thus, computing every permutation requires Θ(n! × n) run time.
Now if you truly understand the algorithm, here’s an extension exercise for you: Design the algorithm for stepping backward to the previous lexicographical permutation. (Spoilers at the bottom.)
Annotated code (Java)
This code can be mechanically translated to a programming language of your choice, with minimal understanding of the algorithm. (Note that in Java, arrays are indexed from 0.)
Example usages
Print all the permutations of (0, 1, 1, 1, 4):
Project Euler #24: Find the millionth (1-based) permutation of (0, 1, 2, 3, 4, 5, 6, 7, 8, 9). My Java solution: p024.java
Project Euler #41: Find the largest prime number whose base-10 digits are a permutation of (1, 2, 3, 4, 5, 6, 7, 8, 9). My Java solution: p041.java
Source code
- Python: nextperm.py
- JavaScript: nextperm.js
- TypeScript: nextperm.ts
- Java: nextperm.java
- C#: nextperm.cs
- C++: nextperm.cpp
- C: nextperm.c
- Rust: nextperm.rs
- Haskell: nextperm.hs (probably suboptimal)
- Mathematica: nextperm.mat.txt
- MATLAB: nextperm.m
License: Nayuki hereby places all code on this page regarding the next permutation algorithm in the public domain. Retaining the credit notice containing the author and URL is encouraged but not required.