There are $n+1C_n = n+1$ combinations possible, with $n!/n= (n-1)! $ orderings possible in each; leading to a total of $(n+1).(n-1)!$.
Let $n= 6.$
2026-03-28 09:33:40.1774690420
Compute the number of distinct actions of cyclic group $C_n$ on a set $X,$ s.t $|X|= n+1.$
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There are two approaches: first take the nine inequivalent actions . Sum up the cases.
Second, take the complement and subtract from 7!. The complementary technique has cases:
a. $1\rightarrow(1234), $
b $1\rightarrow(1234)(56), $
c. $1\rightarrow(1234)(567), $
d. $1\rightarrow(12345)(67), $
e. $1\rightarrow(12345). $
f. $1\rightarrow (1234567).$
Can subtract the sum of above six cases, from $7!$ to get actual value.
There are two techniques used:
First, is to sum up cases, as shown in comment by @StinkingBishop, let us call it : first way.
Second, is the usual approach - called : direct approach.
The complementary technique is easy as the computation is easier, using the first way.
The unwanted set of inequivalent actions is impossible to arrive in result.
The sum of complementary actions is:
a+b+c. $7C4.(4!/4).3!= 35.36= 1260.$
d+e. $7C5.(5!/5).2!=21.24.2=1008.$
f. $7C7.(7!/7)= 6!= 720.$
The sum is: $2268+720= 2988.$
This leaves after subtraction from $7!= 5040$ a value of $2052$.
While the direct approach consists of taking the sum of the below cases:
1.$1\rightarrow e, $
2. $1\rightarrow(123456), $
3. $1\rightarrow(12), $
4. $1\rightarrow(12)(34), $
5. $1\rightarrow(12)(34)(56),$
6. $1\rightarrow(123),$
7. $1\rightarrow(123)(456),$
8. $1\rightarrow(12)(345), $
9. $1\rightarrow(12)(345)(67).$
But, the number of cases here are more, else unwanted actions would be counted; as well as double-counting can arise.
By the direct approach, need find the different cases:
$841+21+ 210+350+630= 862+560+630= 862+1190= 2052.$
Taking the approach of counting (first way) would have reduced number of cases:
1.$1\rightarrow e, $
2. $1\rightarrow(123456), $
3'. $1\rightarrow(12), $
4'. $1\rightarrow(123),$
And count given for cases 3', 4' by : $7C2.5!, 7C3.2.4!= 21.120 + 70.24 = 2520 + 1680.$
The total number of actions are: $1+840+2520+1680= 5041.$
Extra values are: $5041 - 2052= 2989$.
The extra values are due to unwanted inequivalent actions, and double counting.
Unwanted inequivalent actions:
r. $1\rightarrow (12)(3456): 7C2.5C4.3!= 21.5.6= 630,$
s. $1\rightarrow (12)(34567): 7C2.5C5.4!= 21.24= 504,$
t. $1\rightarrow (123)(4567): 7C3.4C4.2.3!= 35.12= 420,$
Double counting :
u. $1\rightarrow (12)(345): 7C2.5C3.2= 21.20= 420,$
v. $1\rightarrow (12)(345)(67): 7C2.5C3= 210,$
The rest values are :
$2989-(630+504+840+210= 630+1344+210= 1974+210= 2184)= 805.$
Some must be accounted by lack of division, for cycles of the same type, as below:
w. $1\rightarrow (12)(34): 105$ extra values,
x. $1\rightarrow (12)(34)(56): 7C2.5C2.3C2 = (21.10.3).5/6= 525$ extra values,
y. $1\rightarrow (123)(456): 7C3.4C3.(3/4)= 35.3= 105$ extra values,
Still have left to account for $70=(805-(105+525+105)= 805-(210+525)= 805-(735))$ more values, and don't know which actions to seek for these.
But, the conclusion is clear: first way works only for complementary approach.