Unizor - Combinatorics - Permutations
- Duration: 18:10
- Updated: 12 May 2014
Combinatorics is a much newer part of mathematics than such classical subjects as Geometry or Algebra. Very important stimulus to its development was the Theory of Probabilities, which is the next subject in this course. Later on, Game Theory and contemporary Computer Science were the other fields of application of the combinatorics.
The subject of Permutations is calculating the number of possibilities of putting certain number of objects in certain order. Examples are numerous. For instance, you have to visit 3 different places A, B and C. The order in which you visit them can be ABC, ACB, BAC, BCA, CAB and CBA. Actually, these 6 different ways to visit 3 places are the only possible ones, there are no other ways of putting them in the order of visiting. In many cases it's very important to know how many possibilities to do something like this visiting exists. That's the task of calculating the permutations.
The simplest form of permutations is putting N objects in some order. For example, we have numbers from 1 to N and want to write them down in some order. We can write them in ascending order, in descending order or in many other types of order. But what is the number of different ways we can write them down?
Ordering of N objects assumes that we have to choose the object #1, then, among other objects we have to choose the object #2, then among whatever is left we have to choose the object #3 etc. Continuing this process to the end, we finally get to object #N.
For #1 object we have N candidates. With each of them there remain N−1 candidates for #2 spot. So, we have N·(N−1) choices for the first 2 places. With each of them there remain N−2 candidates for #3 spot, which brings the total number of variants for the first 3 places to N·(N−1)·(N−2). Continuing this process to all N places, we come up with the total number of variants to position N objects in some order equal to N·(N−1)·(N−2)·...·2·1.
This quantity is usually written in the form N! and pronounced "N factorial".
At this point we'd like to suggest certain criticism to many textbooks that explain this more or less in the same way. The explanation above is just the explanation, not a proof. So, to stop at this point, as many textbooks do, cannot be considered a truly mathematical approach. We have to prove it, and that's exactly what we are going to do next.
Proof
We are going to prove by induction that the number of permutations of N different objects equals to
N·(N−1)·(N−2)·...·2·1 = N!
(that is, N factorial)
Induction step 1. The formula is obviously correct for N=1 because it produces the result 1 and there is only one way to position a single object in some order.
Induction step 2. Assume that the formula is correct for N=K, that is that the number of permutations of K objects is K!.
Induction step 3. Consider we have N=K+1 objects. Each permutation (the way of ordering) of these objects involves positioning of the first K objects in some way (and there are K! of these ways, as we assumed on step 2) and placing the (K+1)th object somewhere among these first K objects. The (K+1)th object can stand before the first, between the first and the second, between the second and the third,..., between (K−1)th and Kth objects and, finally, after the Kth object. This constitutes K+1 different positions for the (K+1)th object. Therefore, for any permutation of the first K objects there are K+1 different permutations of the whole set of K+1 objects. Hence, the total number of permutations of K+1 objects equals to K! multiplied by (K+1).
But K! · (K+1) = (K+1)!, as follows from the definition of the "factorial" as a product of all natural numbers from 1 to a given number. That proves that the formula retains its form when we move from N=K objects to N=K+1 objects.
This completes the proof.
We will use the symbol P(N) to designate the number of permutations of N objects. We, therefore, have proved that
P(N) = N!
http://wn.com/Unizor_-_Combinatorics_-_Permutations
Combinatorics is a much newer part of mathematics than such classical subjects as Geometry or Algebra. Very important stimulus to its development was the Theory of Probabilities, which is the next subject in this course. Later on, Game Theory and contemporary Computer Science were the other fields of application of the combinatorics.
The subject of Permutations is calculating the number of possibilities of putting certain number of objects in certain order. Examples are numerous. For instance, you have to visit 3 different places A, B and C. The order in which you visit them can be ABC, ACB, BAC, BCA, CAB and CBA. Actually, these 6 different ways to visit 3 places are the only possible ones, there are no other ways of putting them in the order of visiting. In many cases it's very important to know how many possibilities to do something like this visiting exists. That's the task of calculating the permutations.
The simplest form of permutations is putting N objects in some order. For example, we have numbers from 1 to N and want to write them down in some order. We can write them in ascending order, in descending order or in many other types of order. But what is the number of different ways we can write them down?
Ordering of N objects assumes that we have to choose the object #1, then, among other objects we have to choose the object #2, then among whatever is left we have to choose the object #3 etc. Continuing this process to the end, we finally get to object #N.
For #1 object we have N candidates. With each of them there remain N−1 candidates for #2 spot. So, we have N·(N−1) choices for the first 2 places. With each of them there remain N−2 candidates for #3 spot, which brings the total number of variants for the first 3 places to N·(N−1)·(N−2). Continuing this process to all N places, we come up with the total number of variants to position N objects in some order equal to N·(N−1)·(N−2)·...·2·1.
This quantity is usually written in the form N! and pronounced "N factorial".
At this point we'd like to suggest certain criticism to many textbooks that explain this more or less in the same way. The explanation above is just the explanation, not a proof. So, to stop at this point, as many textbooks do, cannot be considered a truly mathematical approach. We have to prove it, and that's exactly what we are going to do next.
Proof
We are going to prove by induction that the number of permutations of N different objects equals to
N·(N−1)·(N−2)·...·2·1 = N!
(that is, N factorial)
Induction step 1. The formula is obviously correct for N=1 because it produces the result 1 and there is only one way to position a single object in some order.
Induction step 2. Assume that the formula is correct for N=K, that is that the number of permutations of K objects is K!.
Induction step 3. Consider we have N=K+1 objects. Each permutation (the way of ordering) of these objects involves positioning of the first K objects in some way (and there are K! of these ways, as we assumed on step 2) and placing the (K+1)th object somewhere among these first K objects. The (K+1)th object can stand before the first, between the first and the second, between the second and the third,..., between (K−1)th and Kth objects and, finally, after the Kth object. This constitutes K+1 different positions for the (K+1)th object. Therefore, for any permutation of the first K objects there are K+1 different permutations of the whole set of K+1 objects. Hence, the total number of permutations of K+1 objects equals to K! multiplied by (K+1).
But K! · (K+1) = (K+1)!, as follows from the definition of the "factorial" as a product of all natural numbers from 1 to a given number. That proves that the formula retains its form when we move from N=K objects to N=K+1 objects.
This completes the proof.
We will use the symbol P(N) to designate the number of permutations of N objects. We, therefore, have proved that
P(N) = N!
- published: 12 May 2014
- views: 1