Determine the order of a subset of S_n

  • Thread starter Thread starter minor_embedding
  • Start date Start date
Click For Summary
SUMMARY

The discussion focuses on determining the order of a subset of bijective functions from the set S = {a,b,c,d} where no element maps to itself. The total number of bijective functions is calculated as 4!, and the specific subset is derived by excluding functions where any element maps to itself, leading to the conclusion that there are 9 valid functions. This type of permutation is identified as a 'derangement', which is a well-established concept in combinatorial mathematics.

PREREQUISITES
  • Understanding of bijective functions and permutations
  • Familiarity with combinatorial notation, specifically binomial coefficients
  • Knowledge of derangements and their properties
  • Basic principles of set theory
NEXT STEPS
  • Research the concept of derangements in combinatorial mathematics
  • Explore the application of the inclusion-exclusion principle in counting problems
  • Learn about advanced permutations and their applications in algorithm design
  • Study the properties of bijective functions in relation to group theory
USEFUL FOR

Mathematicians, computer scientists, and students studying combinatorics or discrete mathematics who are interested in understanding permutations and their applications.

minor_embedding
Messages
4
Reaction score
0

Homework Statement


We have a set S = {a,b,c,d} and a set X such that X = {##f:S \rightarrow S| ## f is bijective and ## f(x) \neq x## for each ##x \in S##}. What is the order of the set?

2. Attempt at Solution/My Reasoning

The order of the set of all bijective functions is simply the 4!. There is one element of set of all bijective functions such that f(x) = x, so we subtract 1 from the order of that set.

Now we keep one element set equal to itself and find all functions such that only that element returns itself. Then you have 3 spots in which the other elements can be mapped to. However, you don't want an element to be mapped to itself so you only have 2 ways in which you can complete this. So you have ## {4 \choose 3} \cdot 2!## ways in which you can do this. Where ##{4 \choose 3}## is the number of ways you can choose that one element.

Next we keep two elements equal to themselves and find all functions where the other two are not. So we have only 1 way in which they can be mapped so then we have ##{4 \choose 2} \cdot 1!## functions such that this is true.

In total we should have 9 functions that satisfy the requirements of X. Is my way of thinking about this right? How can we generally determine the order of this subset of functions?
 
Physics news on Phys.org
Yes, it's correct. Permutations like this are called 'derangements'. You can Google that if you want the general case.
 

Similar threads

Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K