### Proof for advance permutation

 PON 1 2 3 4 5 PON 1 2 3 4 5 PON 1 2 3 4 5 PON 1 2 3 4 5 PON 1 2 3 4 5 1 a b c d e 25 a b c e d 49 a b e c d 73 a e b c d 97 e a b c d 2 b a c d e 26 b a c e d 50 b a e c d 74 b e a c d 98 e b a c d 3 a c b d e 27 a c b e d 51 a c e b d 75 a e c b d 99 e a c b d 4 b c a d e 28 b c a e d 52 b c e a d 76 b e c a d 100 e b c a d 5 c a b d e 29 c a b e d 53 c a e b d 77 c e a b d 101 e c a b d 6 c b a d e 30 c b a e d 54 c b e a d 78 c e b a d 102 e c b a d 7 a b d c e 31 a b d e c 55 a b e d c 79 a e b d c 103 e a b d c 8 b a d c e 32 b a d e c 56 b a e d c 80 b e a d c 104 e b a d c 9 a c d b e 33 a c d e b 57 a c e d b 81 a e c d b 105 e a c d b 10 b c d a e 34 b c d e a 58 b c e d a 82 b e c d a 106 e b c d a 11 c a d b e 35 c a d e b 59 c a e d b 83 c e a d b 107 e c a d b 12 c b d a e 36 c b d e a 60 c b e d a 84 c e b d a 108 e c b d a 13 a d b c e 37 a d b e c 61 a d e b c 85 a e d b c 109 e a d b c 14 b d a c e 38 b d a e c 62 b d e a c 86 b e d a c 110 e b d a c 15 a d c b e 39 a d c e b 63 a d e c b 87 a e d c b 111 e a d c b 16 b d c a e 40 b d c e a 64 b d e c a 88 b e d c a 112 e b d c a 17 c d a b e 41 c d a e b 65 c d e a b 89 c e d a b 113 e c d a b 18 c d b a e 42 c d b e a 66 c d e b a 90 c e d b a 114 e c d b a 19 d a b c e 43 d a b e c 67 d a e b c 91 d e a b c 115 e d a b c 20 d b a c e 44 d b a e c 68 d b e a c 92 d e b a c 116 e d b a c 21 d a c b e 45 d a c e b 69 d a e c b 93 d e a c b 117 e d a c b 22 d b c a e 46 d b c e a 70 d b e c a 94 d e b c a 118 e d b c a 23 d c a b e 47 d c a e b 71 d c e a b 95 d e c a b 119 e d c a b 24 d c b a e 48 d c b e a 72 d c e b a 96 d e c b a 120 e d c b a

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(Permutation Order Number) 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 + 0 + 0 + 4 + 12 + 72 = 89

This is the proof of the PON formula. (object into number)

Finding the arrangement of the given number of the permutation

Suppose a permutation order number is given let it be  89 and let the number of objects to be permuted be 5.

We know the formula of the converse PON (Number into Object) 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 2nd 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 1st 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 1st 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 2nd last object. (actual position = r-1)

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

.

.

.

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

r) (a.n.p.p) for 1st 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 1st. 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 1st 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 2nd last object 3rd last object till the 1st 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. x = 89

(r)

according to formula for the last position object dividing x by (r-1)!

r = number of objects should be known, let r =5

so.

89/(5-1)! = 3.708

Which is a decimal number

Then according to case 1 select the number which is before the decimal = 3

Multiplying (r-1)! To the selected number = 3x (r-1)! = 3x4! = 72

The product obtained is called as applied number of particular position or  (a.n.p.p)r

Therefore (a.n.p.p)5 = 72

72 is know as reducer and 89=x is known as the number to be reduced.

Subtracting (a.n.p.p)5=72  with 89 as mentioned in the formula we get. 89-72 = 17 =y

17 is the new number to be reduced.  All the remaining position of objects will be within the number 17

(r-1)

According to formula for the second last object dividing y by (r-2)!

17/ (5-2)! = 17/ 3! = 2.8

Then according to case 1 selecting the number which is before the decimal =2

Multiplying the (r-2)! To the selected number we get (5-2)!x2 = 3!x2 = 6x2 =12

The product obtained is known as applied number of particular position or (a.n.p.p)(r-1)

Therefore (a.n.p.p)(4) = 12

12 is know as reducer and 17=y is known as the number to be reduced.

Subtracting (a.n.p.p)(4) =12 with 17 as it is mentioned in the formula we get     17-12 = 5

5 is known as the new number to be reduced. Let 5=z.

(r-2)

According to formula for 3rd last object dividing z by (r-3)!

5/(r-3)! = 5/2! = 2.5

Then according to case 1 selecting the number which is before the decimal =2

Multiplying the (r-3)! To the selected number we get (5-3)!x2 =2!x2 = 4

The product obtained is known as applied number of particular position or  (a.n.p.p)(r-2)

Therefore  (a.n.p.p)(3) = 4

4 is known as reducer and 5=z is known as the number to be reduced

Subtracting (a.n.p.p)(3) =4 with 5 as it is mentioned in the formula we get  5-4 =1

1 is known as the new number to be reduced. Let 1= l

(r-3)

According to the formula for the 4th last object dividing l by (r-4)!

1/(r-4)! = 1/(5-4)! = 1/1! = 1

Then according to case 2 selecting the number (1-1) =0

Multiplying the (r-4)! To the selected number we get (5-4)!x0 = 1!x0 = 0

The product obtained is known as applied number of particular position or  (a.n.p.p)(r-3)

Therefore (a.n.p.p)(2) = 0

0 is known as reducer and 1=l is known as the number to be reduced

Subtracting (a.n.p.p)(2) = 0 with 1 as it is mentioned in the formula we get 1-0=1

1 is known as the new number to be reduced. Let 1 = k

(r-4)

According to the formula for the 5th last object dividing k by (r-5)!

1/(r-5)! = 1/(5-5)! = 1/1! = 1

Then according to case 2 selecting the number (1-1) = 0

Multiplying (5-5)! To the selected number we get (5-5)!x1 = 0!x0=0

The product obtained is known as the applied number of particular position or (a.n.p.p)(r-4)

Therefore (a.n.p.p)(1)=0

0 is known as reducer and 1=k is known as the number to be reduced

Subtracting (a.n.p.p)(2) = 0 with 1 as it is mentioned in the formula we get 1-0=1

1 is known as the new number to be reduced. Let 1 = j

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

 Actual position of object a.n.p.p ↔number r     = 5 (a.n.p.p)5 72 r-1 = 4 (a.n.p.p)4 12 r-2 = 3 (a.n.p.p)3 04 r-3 = 2 (a.n.p.p)2 00 r-4 = 1 (a.n.p.p)5 00

The linear position of particular object with respect to previous position object will be

(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)!

Therefore the linear position of the last object (5) = 5 – 72/ (5-1)!

5- 72/4! = 5 – 72/24 = 2

Similarly the linear position of the second last object (4) = 4 – 12/ (4-1)!

4- 12/(3)! = 4- 12/6 =2

Similarly the linear position of the 3rd last object (3) = 3- 4/(3-1)!

3- 4/(2)! = 3 – 4/2 = 1

Similarly the linear position of 4th last object or the 2nd object (2) = 2- 0/ (2-1)!

2-0/1! = 2-0/1 = 2-0 = 2

Similarly the linear position of 5th last object or the 1st object (1) = 1 – 0/ (1-1)!

1-0/0! = 1- 0/1 = 1

After finding the linear position for all the objects from last to 1st. 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 1st 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 2nd last object 3rd last object till the 1st object on the basis of linear position of current particular object with respect to previous position object on the unfilled blank space.

This procedure is shown below.

Arranging the blank space according to the linear position of object  we get

__      __      __      __      __

1        2        3        4        5

We know the linear position of the last position object = 2 placing the last position object e on the blank space we get

e

__      __      __      __      __

1        2        3         4        5

While placing the second last object only unfilled space should be taken into consideration. Placing the second last  d object on the basis of its linear position=2 we get

e        d

__      __      __      __      __

1                  2        3        4

While placing the third last object only unfilled space should be taken into consideration. Placing the third last object c on the basis of its linear position= 1 we get.

c        e        d

__      __      __      __      __

1                           2         3

While placing the forth last object or 2nd object only unfilled space should be taken into consideration. Placing the forth last object b on the basis of its linear position= 1 we get.

c        e        d                   b

__      __      __      __      __

1        2

While placing the fifth last object or 1st object only unfilled space should be taken into consideration. Placing the fifth last object a on the basis of its linear position= 1 we get.

c        e        d        a         b

__      __      __      __      __

1

So the required arrangement of object is

c   e    d    a    b

Since this converse is same as that of the arrangement of object the formula is true.

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