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!