Calculate how many PERMUTATION possible

[Sh3nmue]

Hello Me Excel... I really need help again here...
I want to calculate for how many permutation possible for number in cell..

just for example:
in collumn A have list of number or letters
column B is permut by 6
column C is permut by 5 and etc...

	A	B	C	D	E	F
1	Number/Letters	Permute 6	Permute 5	permute 4	Permute 3	Permute 2
2	abcd			24	24	12
3	aabc			12	12	7
4	aaab			4	4	3
5	aabb			6	6	4
6	abcde		120	120	60	20
7	aabcd			60	33	13
8	aabbc			30	18	8
9	abcdef	720	720	360	120	30
10	aabcde
11	aaabcd
12	aabbcd

<colgroup><col style="width: 25pxpx"><col><col><col><col><col><col></colgroup><thead>
</thead><tbody>
</tbody>

Sheet1​

all the value in the table above is calculated manually and it wasting my time an paper,, LoL,, because i have to write down all possible permutation and count it...
can someone help me with this...
Thank You

 

[Sh3nmue]

Bump

 

[StephenCrump]

For what it's worth, the more rigorous mathematical approach to the five letter MISSISSIPPI permutation problem says that given one letter (M) with one occurrence, one letter with two occurrences, and two with four occurrences, the answer is the coefficient of x^5 in:

5! (1 + x) (1+ x + x^2/2!) (1 + x + x^2/2! + x^3/3! + x^4/4!)^2

 

[shg]

Generating function -- I like it!

5x^11/48 + 55x^10/48 + 85x^9/12 + 745x^8/24 + 310x^7/3 + 805x^6/3 + 550x^5 + 880x^4 + 1060x^3 + 900x^2 + 480x + 120

 

[shg]

Oops -- something ran off the rail:

	A​	B​	C​	D​
1​		11​	34,650​	C1: =PRODUCT(C3:C14)
2​	Letter​	Count​	Combinations​	​
3​	M​	1​	11​	C3 and down: =COMBIN(11-SUM(B$2:B2), B3)
4​	I​	4​	210​
5​	S​	4​	15​
6​	P​	2​	1​

 

[StephenCrump]

5x^11/48 + 55x^10/48 + 85x^9/12 + 745x^8/24 + 310x^7/3 + 805x^6/3 + 550x^5 + 880x^4 + 1060x^3 + 900x^2 + 480x + 120

Did you generate this in Excel?

Because the same approach can be used to answer the questions in #20.

 

[shg]

No, at Wolfram|Alpha Examples: Polynomials

But apparently incorrectly.

 

[StephenCrump]

Thanks. I also found this site which was useful: Expand and Simplify Polynomials Calculator

For the MISSISSIPPI problem, the first few coefficients (for 1, x, x^2 ... etc) were

1, 4, 15/2, 53/6, 22/3, 55/12

So the permutation counts are:
1: 4 x 1! = 4
2: 15/2 x 2! = 15
3: 53/6 x 3! = 53
4: 22/3 x 4! = 176
5: 55/12 x 5! = 550 ....

For the 11223 problem above, we need to expand (1+x)(1+x+x^2/2)^2
for which the coefficients are:

1, 3, 4, 3, 5/4, 1/4

So permutation counts are:
1: 3 x 1! = 3
2: 4 x 2! = 8
3: 3 x 3! = 18
4: 5/4 x 4! = 30
5: 1/4 x 5! = 30.

Unless there's a smarter way to calculate coefficients(?) it shouldn't be too difficult to render either in Excel or VBA. But unless there's a smarter way, I'm leaning towards brute-force, i.e. VBA to calculate all permutations and use Excel to eliminate duplicates in the resulting list.

 

[Sh3nmue]

Thanks. I also found this site which was useful: Expand and Simplify Polynomials Calculator

For the MISSISSIPPI problem, the first few coefficients (for 1, x, x^2 ... etc) were

1, 4, 15/2, 53/6, 22/3, 55/12

So the permutation counts are:
1: 4 x 1! = 4
2: 15/2 x 2! = 15
3: 53/6 x 3! = 53
4: 22/3 x 4! = 176
5: 55/12 x 5! = 550 ....

For the 11223 problem above, we need to expand (1+x)(1+x+x^2/2)^2
for which the coefficients are:

1, 3, 4, 3, 5/4, 1/4

So permutation counts are:
1: 3 x 1! = 3
2: 4 x 2! = 8
3: 3 x 3! = 18
4: 5/4 x 4! = 30
5: 1/4 x 5! = 30.

Unless there's a smarter way to calculate coefficients(?) it shouldn't be too difficult to render either in Excel or VBA. But unless there's a smarter way, I'm leaning towards brute-force, i.e. VBA to calculate all permutations and use Excel to eliminate duplicates in the resulting list.

OK I'm so dumb in this...
can u tell me again how did u find this (1, 3, 4, 3, 5/4, 1/4 ) and this (1, 4, 15/2, 53/6, 22/3, 55/12)

 

[Sh3nmue]

Guys I found something... excel vba to generate and count permutation and combination... maybe u all wanna take a look,, I hope this can be a useful clues..
https://app.box.com/s/2irz4lp2u3hm4i46r01i

 

[StephenCrump]

can u tell me again how did u find this (1, 3, 4, 3, 5/4, 1/4 ) and this (1, 4, 15/2, 53/6, 22/3, 55/12)

For the first one, in this website Expand and Simplify Polynomials Calculator in the "Input Polynomial" box, enter: (1+x)(1+x+x^2/2)^2 and hit the "Compute" button. It will expand the polynomial to: 1 + 3x + 4x^2 + 3x^3 + (5/4)x^4 + (1/4)x^5, ie coefficients 1, 3, 4, 3, 5/4 and 1/4.

Guys I found something... excel vba to generate and count permutation and combination... maybe u all wanna take a look,, I hope this can be a useful clues..
https://app.box.com/s/2irz4lp2u3hm4i46r01i

Nice find! Another VBA method that gives you exactly what you're looking for.