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- An inversion of a permutation σ is a pair (i,j) of positions where the entries of a permutation are in the opposite order: i σj. So a descent is just an inversion at two adjacent positions. For example, the permutation σ = 23154 has three inversions: (1,3), (2,3), (4,5), for the pairs of entries (2,1), (3,1), (5,4). Sometimes an inversion is defined as the pair of values.
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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:
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.
Result
| Permutations, nPr = |
| = | 30 |
| Combinations, nCr = |
| = | 15 |
Permutations and combinations are part of a branch of mathematics called combinatorics, which involves studying finite, discrete structures. Permutations are specific selections of elements within a set where the order in which the elements are arranged is important, while combinations involve the selection of elements without regard for order. A typical combination lock for example, should technically be called a permutation lock by mathematical standards, since the order of the numbers entered is important; 1-2-9 is not the same as 2-9-1, whereas for a combination, any order of those three numbers would suffice. There are different types of permutations and combinations, but the calculator above only considers the case without replacement, also referred to as without repetition. This means that for the example of the combination lock above, this calculator does not compute the case where the combination lock can have repeated values, for example 3-3-3.
Permutations
The calculator provided computes one of the most typical concepts of permutations where arrangements of a fixed number of elements r, are taken from a given set n. Essentially this can be referred to as r-permutations of n or partial permutations, denoted as nPr, nPr, P(n,r), or P(n,r) among others. In the case of permutations without replacement, all possible ways that elements in a set can be listed in a particular order are considered, but the number of choices reduces each time an element is chosen, rather than a case such as the 'combination' lock, where a value can occur multiple times, such as 3-3-3. For example, in trying to determine the number of ways that a team captain and goal keeper of a soccer team can be picked from a team consisting of 11 members, the team captain and the goal keeper cannot be the same person, and once chosen, must be removed from the set. The letters A through K will represent the 11 different members of the team:
A B C D E F G H I J K 11 members; A is chosen as captain
B C D E F G H I J K 10 members; B is chosen as keeper
As can be seen, the first choice was for A to be captain out of the 11 initial members, but since A cannot be the team captain as well as the goal keeper, A was removed from the set before the second choice of the goal keeper B could be made. The total possibilities if every single member of the team's position were specified would be 11 × 10 × 9 × 8 × 7 × ... × 2 × 1, or 11 factorial, written as 11!. However, since only the team captain and goal keeper being chosen was important in this case, only the first two choices, 11 × 10 = 110 are relevant. As such, the equation for calculating permutations removes the rest of the elements, 9 × 8 × 7 × ... × 2 × 1, or 9!. Thus, the generalized equation for a permutation can be written as:
| nPr = |
|
Or in this case specifically:
| 11P2 = |
| = |
| = 11 × 10 = 110 |
Again, the calculator provided does not calculate permutations with replacement, but for the curious, the equation is provided below:
nPr = nr
Combinations
Combinations are related to permutations in that they are essentially permutations where all the redundancies are removed (as will be described below), since order in a combination is not important. Combinations, like permutations, are denoted in various ways including nCr, nCr, C(n,r), or C(n,r), or most commonly as simply | ( | n | ) |
| r |
, it does not matter whether A and then B or B and then A
are chosen to be strikers in those respective orders, only that they are chosen. The possible number of arrangements for all n people, is simply n!, as described in the permutations section. To determine the number of combinations, it is necessary to remove the redundancies from the total number of permutations (110 from the previous example in the permutations section) by dividing the redundancies, which in this case is 2!. Again, this is because order no longer matters, so the permutation equation needs to be reduced by the number of ways the players can be chosen, A then B or B then A3 Permute 9
, 2, or 2!. This yields the generalized equation for a combination as that for a permutation divided by the number of redundancies, and is typically known as the binomial coefficient:| nCr = |
|
Or in this case specifically:
Finite Math Calculator
| 11C2 = |
| = |
| = 55 |
It makes sense that there are fewer choices for a combination than a permutation, since the redundancies are being removed. Again for the curious, the equation for combinations with replacement is provided below:
10 Permute 3
| nCr = |
|