[Math Talk #18] Secret of Shuffling - Derangements

2년 전


0. What is a Derangement?

In mathematics, permutation is an act of arranging all the members of a set into some sequence or order. A permutation which leaves no element fixed is called a derangement . In many cases, we deal with items (a finite set), each labeled as

So, put another way, a derangement of objects is a permutation of them without fixed points. The number of derangements of obejcts is usually written as .

For example, , since singleton set can not have any derangements. Also , because only derangement is assigning

Let's look at one more example . There are total 2 derangements,

maps to231
maps to312

In general, we need to ask the question:

Of the different permutations of distinct objects, how many leave no object in its original place?

The answer to the question will provide a general formula for as well as

the probability that a permutation of objects is a derangement.

1. Old but Gold Recurrence Relation

Viewing as a sequence , let's find a recurrence relation for . Since we already know the value of and , let's focus on the values of for .

If a permutation

is a derangement, then it must be that , which leaves possibilities for it. Among those, choose

Now there are two possibilities.

1. If , this is nothing but derangement of

and there are exactly of these.

2. If , this is nothing but derangement of


with a restriction . The situation is totally equivalent to derangement of


if is viewed as ; so there are exactly of these.

Since our choice of was arbitrary, summing all those cases gives

How do we solve this?

Clever trick is to divide both sides by so that

Defining gives

so that


2. The other way around


We've established the formula for derangement, but there is a more intuitive way of relating the total number of permutations with .

We can count the total ways by partitioning the ways into disjoint subsets,

where is the set of permutations in which there are exactly fixed points. If there are fixed points, we first choose fixed points among , and arrange leftovers ( elements) to have no fixed points. This gives

so that

This simple looking relationship is indeed very useful, as we will use in further sections.

3. Expected Number of Fixed points

A rather paradoxical situation arise when we calculate the expeced number of fixed points. Suppose that we perform the experiment of matching the initial arrangement with a random permutation of itself. Define a random variable as the number of fixed points in single experiment. Then clearly, can have values

So the expectation is nothing but

In the previous Section, we already obtained the formula for , which is

Now, straightforward calculation gives


by our previous result in Section 3, we get

regardless of .

How should we interpret this weird result? Many of us tend to think the expected number of fixed points will grow as increases. However, the probability of making an element as a fixed point is

inversely proportional to

This will help you understand what's going on more clearly. Consider a random variable such that if is a fixed point, and 0 otherwise. Then the expectation of number of fixed points is

so that cancel out making the expectation constant; 1.

4. Asymptotic Property of derangement

Finally, one can ask that

What is the asymptotic behavior of as gets large?

This is nothing but asking the limit

one can easily show that the limit actually exists, and surprisingly, using the well known fact

gives .

Also one can ask that

What is the asymptotic behavior of as gets large?

Recall that we defined random variable as number of fixed points in single random permutation? Using the definition in Section 2 gives

We already proved that , so that

5. Computer Simulation

5-1. Expected number of fixed point revisited

Here is the MATLAB simulation for accumulated expectation of number of fixed points for various . As the number of simulation grows, you can see the accumulated expectation approaches to unity, regardless of .

5-2. Asymptotic property of Derangement revisited

Now here is the MATLAB simulation for the probability of having derangement in single permutation.

You can see it clearly converges to , oscillating in small region centered .

6. Conclusion

  • Number of derangement of items is equal to

  • The expectated number of fixed points in single random permutation is 1, regardless of .

  • The probability of having derangement converges to .

7. Citations

[1] Image Source Link, By RokerHRO - Own work, CC BY 3.0

[2] Julian Havil, Nonplussed! Mathematical Proof of Implausible Ideas. Chapter 5 page 53

[3] Julian Havil, Nonplussed! Mathematical Proof of Implausible Ideas. Chapter 5 page 56

MATLAB plots are self-made.

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