Permutations Formulae without Repetitions
In this class, We discuss Permutations Formulae without Repetitions.
The reader should have a prior understanding of permutation basics. Click Here.
We take an example and understand the formulae.
First, We need to understand the concept of repetitions.
Given five alphabets: A, B, C, D, and E.
How many different words are possible with the given alphabet?
Here, the condition is not to use the alphabet twice.
Example:
A A B C D is an example using repetitions. A is used twice.
The formulae we discuss in this class are used without repetition.
Example:
Given five alphabets. A, B, C, D, and E.
How many three-letter words are possible using these five alphabets?
Solution:
Five alphabets should be placed in three positions.
_ _ _
The first position can be filled in five possible ways.
The second position can be filled in 4 possible ways.
The third position can be filled in 3 possible ways.
The number of possibilities is 5*4*3
The number of permutations possible is given Npr
Npr = N!/(N-r)!
Here, N is the number of alphabets
r is the number of positions.
5p3 = 5!/(5-3)!
5!/2!
= (5*4*3*2*1)/(2*1)
= 5*4*3
The formulae match our output.
In our last class, we considered an example of five alphabets in five positions.
The formulae is NpN = N!/(N – N)!
= N!