CHAPTER 12. COMBINATIONS AND PERMUTATIONS 12.6
Now suppose you havethese objects:
1; 2; 3
Here is a list of all permutations of all three objects:
1 2 3; 1 3 2; 2 1 3; 2 3 1; 3 1 2; 3 2 1.
Counting Permutations EMCDM
Let S be a set with n objects. Permutations of r objects from this set S refer to sequences of r different
elements of S (where two sequencesare considered differentif they contain the sameelements but
in a different order). Formulae for the number of permutations and combinations are readily available
and important throughout combinatorics.
It is easy to count the number of permutations ofsize r when chosen from a set of size n (with r≤ n).
- Select the first member of the permutation out ofn choices, because there aren distinct elements
in the set. - Next, since one of the n elements has already been used, the second member of the permutation
has (n− 1) elements to choose fromthe remaining set. - The third member of the permutation can be filled in (n− 2) ways since 2 have been used
already. - This pattern continues until there are r members on the permutation. This means thatthe last
member can be filled in (n− (r− 1)) = (n−r + 1) ways. - Summarising, we find that there is a total of
n(n− 1)(n− 2)··· (n−r + 1)different permutations of r objects, taken from a pool of n objects. This number is denoted by
P (n,r) and can be written in factorial notation as:P (n,r) =
n!
(n−r)!.
For example, if we havea total of 5 elements, the integers{1; 2; 3; 4; 5}, how many ways are there for
a permutation of threeelements to be selectedfrom this set? In this case, n = 5 and r = 3. Then,
P (5,3) = 5!/7! = 60!.
See video: VMihe at http://www.everythingmaths.co.zaExample 3: Permutations