Permutation without Repetition

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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