Permutation

Permutation 

Permutation can be defined as an arrangement of object in a definite order.

We will try to find the number of permutation in various different scenario without actually counting. 

Let us start our disscussion with an illustration. 

Using all the letters of the word MATHS how many distinct words with or without meaning can be formed without repetition ? 

We know that you can solve it with simple knowledge of counting principle

Case 1

Let us consider a word formed with only one alphabet 

So there are 5 words

Case 2

2 alphabet 

Let us physically find out the cases.

(MA),(MT),(MH),(MS),(AT),(AH),(AS),(TH),(TS),(HS)

Now we have 10 cases 

Now we should remember that order matters in this case.

For MA we have another case AM

So we have 1 more cases for every case.

2×10=20

Now let us use our intelligence to solve this question more easily. 

M will form arrangement with 4 other alphabet 

Next one will form arrangement with 3 alphabet and so on

So we have    4+3+2+1=10

Its still quite messy 

We are having 5 alphabet and we need to select 2 alphabets.

Now we have two places

In the first place I have 5 options and now I am left with 4 option 

So we have 5×4 cases

This is based on fundamental principle of counting. 

Case 3

Now we need to select three alphabet from 5

5×4×3=60=  5!   =   5!   

                     2!      (5-3)!

Case 4

We need to select 4 alphabet from 5

   5!     = 120

(5-4)!

Case 5

We need to rearrange them

5×4×3×2×1= 5!  = 120

     (5-5)!        1

(0!=1)

Summing up all the cases we arrive at the point of answering the question. 

120+120+60+20+5=325

If you have studied all the cases correctly than from the third case onwards we arrived at a pattern 

A permutation is a choice of r things from a set of n things without replacement and where the order matters.

Now let us try to solve some problems with the formula 

Illustration 

1)Using all the letters of the word PERMUTATION  how many words with or without meaning containing three alphabet can be formed. 

Answer-990

Solution 

We have 11 alphabet and we need to select 3 alphabet 

P(11,3)  = 11!/(11-3)! 

               =11×10×9=990

Permutation with repetition. (Permutation of multi set)

What will happen if we have 4,4,4,4,6,6,7 - digits and we want to find how many different arrangement we can form ?

Some would say 7! As there are 7 objects but writing 444 and 444 is same. Where's 123 and 321 are different. So in case of repeating elements we have to exclude them. 

So the number of permutation formed by 4,4,4,4,6,6,7 taking all at a time , without taking any element more than once would be            7!   

            4!2!1!

Now why I am dividing it by 4! , as 4 is repeating it self 4 times.