Information adapted from the following website:

http://www.algebra.com/algebra/homework/Permutations/change-this-name6021.lesson

This is a good one. WATCH THIS MORE THAN ONCE!!!

Permutation

-A permutation is "a re-arrangement of elements of a set."

-It means a permutation is ONLY interested in re-arranging the elements of the set... Any duplications of the collected elements in different orders is IMPORTANT TO COUNT. A permutation therefore tends to be a large number, because ORDER COUNTS!

Example:
Taking the 4 letters, ABCD, write down all the permutations of 3 of these leters:

ABC BAC CAB DAB
ACB BCA CBA DBA
ACD BCD CBD DBC

-There are 24 permutations.

-The order matters, since ABC is different to ACB and different to BCA and different to CAB etc. Permutations see these as all different answers.

Combination

-A combination is "one or more elements selected from a set without regard to the order"

-The "without regard" means that the collection matters rather than order in combinations, so in the above example, the fact we ABC, ACB, BAC, BCA, CAB, CBA... for combinations, these are all 1 combination of letters A, B and C. ORDER DOES NOT MATTER.

-So, questions concerning picking a team of 5 people from a squad of 11... you would need combinations, since it is having "Bert, Ernie, Fred, Bill and Bob" that matters, not the fact that you have so many different permutations of these 5 people.

Example:
Taking the 4 letters, ABCD, write down all the combinations of 3 of these letters:

ABC ABD ACD BCD

-There are just 4 combinations. You cannot pick any other 3 letters from ABCD, that is not part of the above 4 combinations. It is enlightening to see the letter missing in each: in order we have "no D", "no C", "no B" and finally "no A"... this sometimes helps you to "see" all the possible answers.

To calculate combinations, it is a 2-stage process:
1. You calculate all the equivalent permutations first.
2. You then correct this list by cutting out any duplicates.

As you can see from this, combinations are a subset (smaller element) of Permutations.

Mathematics of Permutations...the order of things DOES matter!
To find the number of permutations of r elements from a set of n, the formula is:

 P%28n%2Cr%29+=+%28n%21%29%2F%28n-r%29%21

Where "P" is the "permutation" and "r" would be the "things" you are arranging and "n" is the total number of things that could be arranged.

So, the above example taking the letters A, B, C, D and writing down all the combinations of 3 letters would be
 P%284%2C3%29+=+%284%21%29%2F%284-3%29%21

 P%284%2C3%29+=+%284%21%29%2F1%21

 P%284%2C3%29+=+%284%2A3%2A2%2A1%29%2F1

 P%284%2C3%29+=+24

Mathematics of Combinations...the order of things DOES NOT matter!
To find the number of combinations of k elements from a set of n, the formula is:

 C%28n%2Ck%29+=+%28n%21%29%2F%28k%21%28n-k%29%21%29

So, the above example would be
 C%284%2C3%29+=+%284%21%29%2F%283%21%284-3%29%21%29

Where "C" is the "combination" and "k" would be the "things" you are arranging and "n" is the total number of things that could be arranged.
 C%284%2C3%29+=+%284%21%29%2F%283%21%281%29%21%29

 C%284%2C3%29+=+%284%2A3%2A2%2A1%29%2F%283%2A2%2A1%2A1%29

 C%284%2C3%29+=+4

EXAMPLE:
Q: From a squad of 16 members, find the total number of different arrangements of 9 players.

A: Now, the question, as usual has no mention of "permutation" or "combination", so we have to figure it out. Having read the above explanations now, hopefully you will appreciate that the question is one about combinations.

It is asking "find the number of combinations of 9 players from a squad of 16.

solution is:
 C%2816%2C9%29+=+%2816%21%29%2F%289%21%2816-9%29%21%29

 C%2816%2C9%29+=+%2816%21%29%2F%289%217%21%29

use a calculator or work it out long handed:
 C%2816%2C9%29+=+%2816%2A15%2A14%2A13%2A12%2A11%2A10%2A9%2A8%2A7%2A6%2A5%2A4%2A3%2A2%2A1%29%2F%289%2A8%2A7%2A6%2A5%2A4%2A3%2A2%2A1%2A7%2A6%2A5%2A4%2A3%2A2%2A1%29

with cancelling, we have
 C%2816%2C9%29+=+%2816%2A15%2A14%2A13%2A12%2A11%2A10%29%2F%287%2A6%2A5%2A4%2A3%2A2%2A1%29

with further cancelling, we have
 C%2816%2C9%29+=+%2816%2A15%2A14%2A13%2A12%2A11%29%2F%287%2A6%2A4%2A3%29

 C%2816%2C9%29+=+%2816%2A15%2A14%2A13%2A11%29%2F%287%2A6%29

 C%2816%2C9%29+=+%2816%2A15%2A2%2A13%2A11%29%2F%286%29

 C%2816%2C9%29+=+%2816%2A15%2A13%2A11%29%2F%283%29

 C%2816%2C9%29+=+%2816%2A5%2A13%2A11%29

which gives C(16,9) = 11440.

Explained differently:

Permutations and Combinations: What is Permutation?

“What are the possible ways you can arrange a group of things” is the simplest way to define permutation.
For example, you have three different playing cards (say, A, B and C). How many ways can you arrange them? Let’s see the solutions:
[A B C], [A C B], [B C A], [B A C], [C A B], [C B A]
There are six possible ways to arrange three different cards.
In the example above we have actually made the all possible arrangements and then counted the number of arrangements (six). However, if the number of cards are more like five hundreds or thousands then it will be very difficult to make all possible arrangements and find out the number of arrangements. Hence in permutation theory we use formulas to directly find out the number of all possible arrangements. With permutation we concentrate more on finding out the number of possible arrangements rather than the arrangements itself.

Permutation and Combinations: What About Repetition?

In the previous playing card example you might have noticed that in any of the arrangement no cards have been used more than once i.e. no repetition of cards. But if repetitions are allowed then what would have been the number of arrangements? Let’s see.
• You have to select three cards out of three; the first card can be selected by three possible ways.
• Again for each selection of first card second card could be selected by three possible ways.
• So for all the selections of first card second card could be selected by 3 X 3 = 9 ways.
• Similarly the total number of possible selections of all the three cards should be
3 X 3 X3 = 27
In line to the above discussions, if you have n items and you have to make sets of r items then total number of possible arrangements (where repetitions of the items are possible) will be:
n X n X ……r times = nr

Permutation where repetitions are not allowed

Let’s again go to the previous example of cards (card A, B, C) to see how permutations need to be calculated when repetitions are not allowed.
• The first card could be selected by 3 possible ways.
• For each selection of the first card second card could be selected by 2 possible ways (since repetitions are not allowed).
• Similarly, for each selections of first and second card third card could be selected by 1 possible way.
• Hence total number of permutations in case of no repetitions is:
3 X 2 X 1 = 6.

If we try to generalize the above discussion for n number of elements, number of possible permutations for all the n items will be:

n x (n-1) x (n-2)……..3x2x1= n!
Now, let’s take the case of selecting only two cards out of three from the already discussed playing card example. Here it goes.
• The first card could be selected again by 3 possible ways.
• For each selections of the first card second card could be selected by 2 ways. So, total numbers of possible permutation for selecting two cards out of three will be:
3x2=6.
If we generalize this discussion in terms of n and r (where, n is the total numbers of items and r is the number of items to be selected for each set), we can derive the formula of permutation as:
n P r = n x (n-1) x (n-2) x (n-3) x (n-r+1)
= n! / (n-r)!