Proving ##\frac{n!}{r(n-r)!}## Distinct r-Cycles in $S_n$

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Therefore, we take the number of permutations of r-1 objects, which is (r-1)!, and multiply it by the number of ways to choose the remaining n-r objects from the remaining n-r objects, which is ##\frac{n!}{(n-r)!}##. This gives us ##\frac{n!}{r(n-r)!}## distinct r-cycles.In summary, in $S_n$, there are ##\frac{n!}{r(n-r)!}## distinct r-cycles, which can be shown by fixing one object and considering the number of permutations of the remaining objects. This results in ##\frac{n!}{r(n-r)!}## distinct r-cycles.
  • #1
Lee33
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Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}## but I don't know how they got ##\frac{n!}{r(n-r)!}##?
 
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  • #2
Lee33 said:

Homework Statement



In $S_n$, prove that there are ##\frac{n!}{r(n-r)!}## distinct r-cycles.

2. The attempt at a solution

I know there are n choose r ways to permute r out of n cycles thus ##\frac{n!}{r!(n-r)!}##

Two r-cycles are the same if their cycle notations are cyclic permutations of each other. Having chosen our objects, we can avoid such multiple counting by fixing one object to appear the beginning of all the r-cycles. Each permutation of the remaining [itex]r-1[/itex] objects will then generate a distinct r-cycle.
 

FAQ: Proving ##\frac{n!}{r(n-r)!}## Distinct r-Cycles in $S_n$

1. What is the purpose of proving ##\frac{n!}{r(n-r)!}## distinct r-cycles in $S_n$?

The purpose of proving this formula is to understand the number of ways in which r distinct elements can be arranged in a group of n elements, where order matters. This is important in various fields of mathematics, such as combinatorics and probability.

2. How is the formula ##\frac{n!}{r(n-r)!}## derived?

The formula can be derived using the concept of permutations, where the number of ways to arrange r distinct elements in a group of n elements is given by nPr = n(n-1)(n-2)...(n-r+1). This can be simplified to ##\frac{n!}{(n-r)!}##, but this includes arrangements where the order of the r elements does not matter. To account for this, we divide by r!, resulting in the formula ##\frac{n!}{r(n-r)!}##.

3. Can you provide an example of using this formula?

Sure. Let's say we have a group of 5 people and we want to know how many ways we can choose 3 of them to form a committee. Using the formula, we have ##\frac{5!}{3(5-3)!} = \frac{120}{6} = 20##. This means there are 20 ways to choose 3 people out of 5 to form a committee.

4. How is this formula related to the concept of cycles in permutation groups?

This formula is directly related to the number of possible cycles in a permutation group. It tells us the number of ways we can choose r distinct elements from a group of n elements and arrange them in cycles, where each element appears exactly once in each cycle. This is useful in studying the structure of permutation groups and their properties.

5. Are there any real-world applications of this formula?

Yes, there are many real-world applications of this formula. For example, in coding theory, this formula is used to determine the number of distinct error patterns that can be corrected by a particular error-correcting code. It is also used in cryptography, where it helps in analyzing the strength of various encryption algorithms.

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