Permutations (last question of sheet, yay )

In summary, the conversation discusses a permutation problem involving determining the sign of a permutation and proving a statement involving cycles and arbitrary permutations. The first part is solved to be (-1)^(n/2) for even values of n and (-1)^((n-1)/2) for odd values of n. The second part is still being worked on and any assistance is appreciated.
  • #1
karnten07
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Permutations (last question of sheet, yay!)

1. Homework Statement [/b]

[tex]\eta[/tex]:=
(1 2 ... n-1 n)
(n n-1 ... 2 1)
[tex]\in[/tex]S[tex]_{n}[/tex] for any n[tex]\in[/tex]N
n.b That should be 2 lines all in one large bracket btw
a.) Determine its sign.

b.) Let n [tex]\geq[/tex]1. Let <a1,...,as> [tex]\in[/tex]Sn be a cycle and let [tex]\sigma[/tex][tex]\in[/tex]Sn be arbitrary. Show that

[tex]\sigma\circ[/tex] <a1,...,as> [tex]\circ[/tex][tex]\sigma^{-1}[/tex] = <[tex]\sigma[/tex](a1),...,[tex]\sigma[/tex](as)> in Sn.

Homework Equations





The Attempt at a Solution



I get the sign of the permutation to be (-1)^n/2

I don;t know how to do the second part, any ideas?
 
Last edited:
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  • #2
Actually, i thought i had done the first part, but i haven't because I am stuck on how to show that for negative numbers, i want the n/2 to be taken as the rounded down value. For example if n=7 i want n/2 to be taken as 3. Is there is a simple way to do this for odd numbers but also keep the same form for positive values of n.

Also it should be (-1)^(n-2/2)

So it should be (-1)^(n-2/2) for even numbers of n and (-1)^(n-3/2) for odd values of n, is there a neater way to do this?
 
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  • #3
I have just said for even numbers of n, that is, n/2 [tex]\in[/tex]Z and for odd numbers, that is n/2[tex]\notin[/tex]Z to distinguish between the two cases.

Im still thinking through part b.) so any help is welcomed.
 

What is a permutation?

A permutation is an arrangement of objects in a specific order. It is a way of selecting a subset of objects from a larger set and arranging them in a particular sequence.

How do you calculate the number of permutations?

The number of permutations of a set of n objects is given by n factorial (n!). This means multiplying n by all the numbers that come before it, for example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

What is the formula for finding the number of permutations?

The formula for finding the number of permutations is n! / (n-r)!, where n is the total number of objects and r is the number of objects being selected and arranged.

What is the difference between permutations and combinations?

Permutations take into account the order of the objects being selected and arranged, while combinations do not. This means that in permutations, the order of the objects matters, whereas in combinations, the order does not matter.

What are some real-life applications of permutations?

Permutations are used in various fields such as mathematics, statistics, computer science, and genetics. In real life, they can be applied in areas such as probability, scheduling, and data analysis.

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