Before we proceed, bear in mind that you need the idea of factorial in this topic.
A factorial is the product of all positive integers less than or equal to a given positive integer. We write this using an exclamation mark (!), for example, 5! = 5 × 4 × 3 × 2 × 1.
A permutation is an ordered arrangement of objects. In other words, it's a way of arranging a set of things where the order matters. The number of permutations of n objects taken r at a time (written as nPr) is calculated using a specific formula.
For example, in how many ways can numbers 123 be arranged?
Answer: 6
Solution: It can be arranged as follows:
ABC, ACB, BAC, BCA, CAB & CBA
TYPES OF
PERMUTATION
|1. Permutation without repetition (Linear Permutation)
2. Permutation with Repetition or Replacement
3. Distinguishable Permutation
4. Circular Permutation
PERMUTATION WITHOUT REPETITION
The number of permutations of a set of n objects taken r at a
time, denoted by nPr, is given by
nPr= n!/(n-r)! ;where, r is less than or equal to n
Where:
P is the total number of permutations
n is the total number of objects in the set
r is the total number of choosing objects in the set
Sample Problems: Solve for the following Permutations.
1. 4P2
Given: P=?; n = 4; r = 2
Formula:
nPr = n!/(n-r)!
Solution:
4P2 = 4!/(4-2)!
4P2 = 4!/2!
4P2 = (4*3*2!)/2!
4P2 = 12 Ways
2. 11P3
Given: P=?; n = 4; r = 2
Formula:
nPr = n!/(n-r)!
Solution:
11P3 = 11!/(11-3)!
11P3 = 11!/8!
11P3 = 11*10*9*8!/8!
11P3 = 11!/8!
11P3 = 11*10*9
11P3 = 990 Ways
3. In how many ways can letters of
the set 1, 2, 3, 4, 5} be arranged to
form ordered codes of 3 letters?
Given: P = ?; n = 5; r = 3
Formula: nPr = n!/(n-r)!
Solution:
5P3 = 5!/(5-3)!
5P3 = 5!/(5 -3)!
5P3 = 5!/2!
5P3 = 5*4*3*2!/2!
5P3 = 60 ways
SOLUTION SHORTCUT: Expand the n factorial until the value of r, but without including it, that's already the answer.
1. 4P2 : 4 - 2 = 2 : 4*3 = 12
2 11P3 : 11 - 3 = 8 : 11*10*9 = 990
3. 5P3 : 5 - 3 : 5*4*3 = 60
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