JUNCTION
Updated Jul 9, 2013, 9:26 PM
This is a result of my extensive research on combination and permutation. I have found an consistent and uniform result for combination as well as permutation and written down on a book known as Junction (An art of counting combination and permutation.If you are interested in the study of junction then e-mail me at softlaws4095@gmail.com
Use template

proof for advance permuting


Finding the PON of the given arrangement of the permutation

Suppose and arrangement of the permutation is given let it be as shown below

89

c

e

d

a

b

 

This is one of the arrangement of 5P5 permutation sequence. Its PON is shown in the first Colum.

We have to find by formula the this number 89 towards the corresponding arrangement of object    c   e   d   a   b

We know the formula of PON (object into number)

Formula permuting (object into number)

1+

1)    (Actual Position of 1st object-1)! x (Total number of previous objects in the permutation arrangement which is right side of the 1st object)+

2)    (Actual Position of 2nd object-1)! x (Total number of previous objects in the permutation arrangement which is right side of the 2nd object)+

3)    (Actual Position of 3rd object -1)! x (Total number of previous objects in the permutation arrangement which is right side of the 3rd object) +

.

.

.

.

        r) (Actual Position of last object-1)! X (Total number of previous objects in the permutation arrangement which is right side of the last object)

 

 

 

 

we know the actual positions and linear positions of each objects

c   e   d   a   b

3   5   4   1   2………… (actual position of object)

1   2   3   4   5…………(linear position of object)

We know the number of objects (r) = 5

Substituting the required in the formula of PON

Now considering the first term of PON

1)    (Actual Position of 1st object-1)! x (Total number of previous objects in the permutation arrangement which is right side of the 1st object)+

The actual position of 1st object = 1

Total number of previous objects in the permutation arrangement which is right side of 1st object = 0

Since there is no object previous to a       

Substituting the value we get

1)    (1-1)! x (0) = 0

 

Similarly for second term of PON

2)    (Actual Position of 2nd object-1)! x (Total number of previous objects in the  permutation arrangement which is right side of the 2nd object)+

 

The actual position of b = 2nd object = 2

Total number of previous objects in the permutation arrangement which is right side of 2nd object = 0

c   e   d   a   b

As per the given arrangement of object there is no object right side of b because b is at the end of the arrangement.

Substituting the value we get.

2)    (2-1)! x (0) = 0

 

Similarly for the third term of PON

3)    (Actual Position of 3rd object -1)! x (Total number of previous objects in the permutation arrangement which is right side of the 3rd object) +

The actual position of c = 3rd object = 3

Total number of previous objects in the permutation arrangement which is right side of 3rd object = 2

c   e   d   a   b

Substituting the value we get.

3)    (3 -1)! x (2) = (2)! x (2) = 2 x 2 = 4

 

Similarly for the 4th term of PON

4)    (Actual Position of 4th object -1)! x (Total number of previous objects in the permutation arrangement which is right side of the 4th object) +

The actual position of d = 4th object = 4

Total number of previous objects in the permutation arrangement which is right side of 4th object = 2

c   e   d   a   b

Substituting the value we get

4)  (4 -1)! x (2) = (3)! x (2) = 6 x 2 = 12

Similarly for the 5th term of PON

5)    (Actual Position of 4th object -1)! x (Total number of previous objects in the permutation arrangement which is right side of the 4th object)

The actual position of d = 5th object = 5

Total number of previous objects in the permutation arrangement which is right side of 5th object = 3

c   e   d   a   b

Substituting the value we get

5)        (5 -1)! x (3) =  (4)! x (3) = 24 x 3 = 72

We have found numbers corresponding to each and every term now substituting as per the formula we get.

The formula of PON

1+

1)    (Actual Position of 1st object-1)! x (Total number of previous objects in the permutation arrangement which is right side of the 1st object)+

2)    (Actual Position of 2nd object-1)! x (Total number of previous objects in the permutation arrangement which is right side of the 2nd object)+

3)    (Actual Position of 3rd object -1)! x (Total number of previous objects in the permutation arrangement which is right side of the 3rd object) +

.

.

.

.

        r) (Actual Position of last object-1)! X (Total number of previous objects in the permutation arrangement which is right side of the last object)

= 1 +   + + 4 + 12 + 72 = 89

This is the proof of the PON formula. 

Comments