In: Advanced Math
Any permutation in has either of the order 1,2,3,4,5,or 6.
S.No. |
Cycle notation |
Type even/odd |
order |
In A5 yes/no |
count |
1 |
(abcde) |
even |
5 |
yes |
24 |
2 |
(abcd) |
odd |
4 |
no |
|
3 |
(abc) (de) |
odd |
6 |
no |
|
4 |
(abc) |
even |
3 |
yes |
|
5 |
(ab) (cd) |
even |
2 |
yes |
15 |
6 |
(ab) |
odd |
2 |
no |
10 |
7 |
identity |
even |
1 |
yes |
Out of these the even permutations of order 2 are of the type (ab) (cd) and odd permutations of order 2 are of the type (ab) .
We need to calculate these two types.
The first is even permutations of order 2 are of the type (ab) (cd).
Note that we have 5 letters a,b,c,d,e so to fill the four places we have possibilities, however, out of these 120 possibilities the following 8 types will be assumed as same
Therefore, we need to divide 120 by 8 which gives 15.
Thus in A5 there are 15 elements order 2.
Next the odd permutations of order 2 are of the type (ab), for this simply we choose 2 letters out of 5 in C(5,2)=10 ways.
Thus in S5 there are 15(even)+10(odd)=25 elements order 2.
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