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Permutation Combination(Addition)Formula

View of the formula: by words

First,it need to take 1's digits as absolutely same as N. Let N=1234,necessary to take 1's digits 1111.It could be computed by this method.
1's digits relative to N:
I=10^d-1+10^d-3+....+10^0 where d=digit count of N
For example,N(=1234):
I=10^3+10^2+10^1+10^0
1111
Then,A=value of adding each digit of N
For N=1234
A=1+2+3+4=10
Next,we must evaluate F(factorial)={(d(digit count)-1)}
If N=1234,then F=(d-1)!=3!=6
You'll get final solution Multiply by values of I,A,F(If no digit of N repeated)
Sometimes,digits of N may be repeated.For that case when you occur(e.g,N=1224),you
essentially require to use factorial reducing method.To achieve this,another variable R is used.If there are repetitive numbers in N.The very first found repetitive digits group is represented by r1.The second sets r2 and the third r3 and so on.For example
(1)N=1332[You can see two three-digits are repeated]
So we get only r1=2
(2)N=15253133[See two one-digits,two five-digits,and three-digits are repeated]
r1,r2,r3 triple must be taken.
r1=2.r2=2,r3=3
To be noticed,the length of r1,r2,....rn is as many as the times of repeated digits.
Another variable R is to be used to store the multiplicative value of r1=rr3=0 and R=1
So,the formula goes this way,
(I*A*F) / R

View of the fornula: by mathematical expression

The Nature of this formula is

I * A * F
R

I=10^d-1 + 10^d-2 +... +10^0
A=value of adding each digit of N
R= r1!r2!r3!...(when only repeated digits occured in N)
Where ,
d=digit count of N
r1,r2,r3=number of repetitive digit group occurred in N
the length of r1,r2,...,rn already expressed is as many as the times of repeated digits meaning that rn=occurrences of repeated digits of N
For example, let N=1223
For example, let N=1223
d=4
I=10^3 + 10^2 + 10^1 +10^0=1111; A=8;F=3!=6;R=r1!=2!=2
Substituting them in formula,
(1111 * 8 * 6) / 2.


Proofs of the formula

Without using the formula

Example (1)N=123
123
132
213
231
312
+ 321 =
1332

by using the formula

For three digits
1=111
A=6
F=2
R=1
Solution (111*6*2)=1332


Example(2) N=557
557
575
+ 755 =
1887

I=111
A=17
F=2
R=2
Solution(111*17*2)/2=1887




Example(1) N=1234
1234
1243
1324
1342
. . . .
. . . .
+ 4231=
66660

For four digits
I=1111
A=10
F=6
R-=1
Solution(1111*10*6)/2=66660


Example(2) N=1223
1223
1232
1322
. . . .
. . . .
+ 3221 =
26664

I=1111
A=8
F=6
R=2
Solution(1111*8*6)/2=26664


Example(1) N=12345
12345
12354
12435
. . . . .
. . . . .
+ 54321=
3999960




For five digits
I=11111
A=15
F=24
Solution(11111*15*24)/1=3999960


Example(2) N=23345
23345
23354
23435
. . . . .
. . . . .
+ 54322=
2266644

I=11111
A=17
F=24
R=2
Solution=2266644


Example(3) N=12233
12233
12323
12322
. . . . .
. . . . .
+ 33221=
733326

I=11111
A=11
F=24
R=r1!r2!=2*2=4
Solution=(11111*11*24)/4=733326










Example(4) N=11126
11122
11212
11221
. . . . .
. . . . .
+ 22111=
155554

I=1111
A=7
F=24
R=r1!r2!=6*2=12
Solution(11111*7*24)/12=155554


Example(5) N=11112
11112
11121
11211
. . . . .
. . . . .
+ 21111=
66666

I=11111
A=6
F=24
R=r1!=24=24
Solution=(11111*6*24)/24=66666

















For six digits
Example(10 N=123456
123456
123465
123546
. . . . . .
. . . . . .
+ 654321=
279999720


I=111111
A=21
F=120
R=1
Solution(111111*21*120)/1=279999720