Before proceeding further we make firm with the terms

a  e  d  b   (Object of P-Junction)

1    4    3    2   (actual position of object in the junction)

1    2    3    4   (linear position of object in the junction)

1    5    4    2   (object number)

Formula permuting (number into object)

Let x be the permutation order number

1)    For last object

Divide x by (r-1)!

Case1€

 

When x/(r-1)! = decimal number

Then select the number which is before the decimal.

 

Multiply (r-1)! To the selected number which was before the decimal.

 

The product obtained by multiplication is called acquired number of particular position for last object

 

i.e. (a.n.p.p) _r

 

Subtract (a.n.p.p)_r with x

 

Then we get x- (a.n.p.p)_r =y

x is known as the number to be reduced (a.n.p.p)_r is known as the obtained reducer. y is known as the new number to be reduced.

 

 

Case2€

 

When x/(r-1) = rational number

Then take the number as (rational number-1)

 

Multiply (r-1)! to the number (rational number-1)

The product obtained by multiplication is called acquired number of particular position for last object.

 

I.e. (a.n.p.p)_r

 

Subtract (a.n.p.p)_r with x then we get

 

X-(a.n.p.p)_r =y

 

X is known as the number to be reduced, (a.n.p.p)_r is known as the obtained reducer. y is known as the new number to be reduced.

 

2)    For second last object.

Same procedure is applied to the new number to be reduced.

 

y/(r-2)! = Either (Case1 or Case2)

 

Where y-(a.n.p.p)_(r-1)= z

.

.

.

.

       (r-1) For 2^nd object.

Let the new number to be reduced be j

Dividing j by 1!

Which will be j itself and it will be either 2 or 1 which is a rational number.

When j=2 take the multiplier (j-1)=(2-1)=1 according to case 2

When j=1 take the multiplier (j-1)=(1-1)=0 according to case 2

Multiply 1! By the occurred number to (j-1). The product which occur is the acquired number of particular position for second object which will be 1or 0.

J-(a.n.p.p)₍₂₎=k

Where k=1

(r) For 1^st object.

The new number to be reduced is k

Divide k by 0!

k/0!=1 which is a rational number.

So the multiplier will be (1-1) =0 according to case 2

Multiply 0! By 0 The product obtained will be acquired number of particular position for 1^st object which will be 0

k-(a.n.p.p)₍₁₎ = l = 1

We have found all the (a.n.p.p) for all the objects from 1 to r.

Arranging all the (a.n.p.p) corresponding to particular position as shown we get

1)    (a.n.p.p)_r for last object. (actual position =r)

2)    (a.n.p.p)_(r-1) for 2^nd last object. (actual position = r-1)

3)    (a.n.p.p)_(r-2) for 3^rd last object.  (actual position = r-2)

.

.

.

        r-1) (a.n.p.p) for 2^nd object (actual position =2)

            r) (a.n.p.p) for 1^st object. (actual position =1)

Linear Position compared to previous object will be.

Linear position of objects in the junction compared to previous position objet

We have found the (a.n.p.p) of each and every object now it is the time to place each and every object according to its linear position in the junction.

(Linear position of current particular object with respect to previous position object)=(actual position of object) - (acquired number of particular position)/(actual position of object-1)!

Find all the linear position of current particular object with respect to previous position object.

After finding the linear position as stated above for all the objects from last to 1^st. Now is the time to place the object.

Before placing the object some arrangement should be done.

Arrange the blank equal to the number of objects (r) as shown below.

__      __      __      __      __      __   . . . . . . . . . . . __      __

  1        2        3        4        5        6                           (r-1)     (r)       (linear position)

Fill in the blanks on the basis of the linear position of current particular object with respect to previous position objects. Start placing the object from last object to 1^st object

Place the last position object on the basis of its linear position of current particular object with respect to previous position object.

After placing the last position object it is the time to place the second last position object.

While placing the second last position object only unfilled blank is taken into consideration.

Similarly place the 2^nd last object 3^rd last object till the 1^st object on the basis of linear position of current particular object with respect to previous position object on the unfilled blank space.

This will be the linear position of objects.