= 1 + 0 + 0 + 4 + 12 + 72 = 89This 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 (r1)!
Case1€
When x/(r1)! = decimal number
Then select the number which is
before the decimal.
Multiply (r1)! 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/(r1) = rational number
Then take the number as (rational
number1)
Multiply (r1)! to the number
(rational number1)
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/(r2)! = Either (Case1 or Case2)
Where y(a.n.p.p)_{(r1)}= z
.
.
.
.
(r1) 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 (j1)=(21)=1 according to case 2
When j=1 take the
multiplier (j1)=(11)=0 according to case 2
Multiply 1! By the
occurred number to (j1). 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 (11) =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)_{(r1)} for 2^{nd}
last object. (actual position = r1)
3)
(a.n.p.p)_{(r2)} for 3^{rd}
last object. (actual position = r2)
.
.
.
r1) (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
object1)!
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 (r1) (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.
x = 89
(r)
according to formula
for the last position object dividing x by (r1)!
r = number of objects should be
known, let r =5
so.
89/(51)! = 3.708
Which is a decimal number
Then according to case 1 select the
number which is before the decimal = 3
Multiplying (r1)! To the selected
number = 3x (r1)! = 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. 8972 = 17 =y
17 is the new number to be reduced. All the remaining position of objects will be
within the number 17
(r1)
According to formula
for the second last object dividing y by (r2)!
17/ (52)! = 17/ 3! = 2.8
Then according to case 1 selecting
the number which is before the decimal =2
Multiplying the (r2)! To the
selected number we get (52)!x2 = 3!x2 = 6x2 =12
The product obtained is known as
applied number of particular position or (a.n.p.p)_{(r1)}
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 1712 = 5
5 is known as the new number to be
reduced. Let 5=z.
(r2)
According to
formula for 3^{rd} last object dividing z by
(r3)!
5/(r3)! = 5/2! = 2.5
Then according to case 1 selecting
the number which is before the decimal =2
Multiplying the (r3)! To the
selected number we get (53)!x2 =2!x2 = 4
The product obtained is known as
applied number of particular position or
(a.n.p.p)_{(r2)}
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 54 =1
1 is known as the new number to be
reduced. Let 1= l
(r3)
According to the
formula for the 4^{th} last object dividing l by (r4)!
1/(r4)! = 1/(54)! = 1/1! = 1
Then according to case 2 selecting
the number (11) =0
Multiplying the (r4)! To the selected
number we get (54)!x0 = 1!x0 = 0
The product obtained is known as
applied number of particular position or
(a.n.p.p)_{(r3)}
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 10=1
1 is known as the new number to be
reduced. Let 1 = k
(r4)
According to the
formula for the 5^{th} last object dividing k by (r5)!
1/(r5)! = 1/(55)! = 1/1! = 1
Then according to case 2 selecting
the number (11) = 0
Multiplying (55)! To the selected
number we get (55)!x1 = 0!x0=0
The product obtained is known as the
applied number of particular position or (a.n.p.p)_{(r4)}
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 10=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

r1 = 4

(a.n.p.p)_{4}

12

r2 = 3

(a.n.p.p)_{3}

04

r3 = 2

(a.n.p.p)_{2}

00

r4 = 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
object1)!
Therefore the linear position of the
last object (5) = 5 – 72/ (51)!
5 72/4! = 5 – 72/24 = 2
Similarly the linear position of the
second last object (4) = 4 – 12/ (41)!
4 12/(3)! = 4 12/6 =2
Similarly the linear position of the
3^{rd} last object (3) = 3 4/(31)!
3 4/(2)! = 3 – 4/2 = 1
Similarly the linear position of 4^{th}
last object or the 2^{nd} object (2) = 2 0/ (21)!
20/1! = 20/1 = 20 = 2
Similarly the linear position of 5^{th}
last object or the 1^{st} object (1) = 1 – 0/ (11)!
10/0! = 1 0/1 = 1
After finding the linear position 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 (r1) (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 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
2^{nd} 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 1^{st} 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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