23.5.2. Overall Formula for Derangements

The last question is we want to find out a recurrence relation for the number of derangements of n objects, so just to recap a derangement of n objects is a permutation of those n objects such that none of the objects is at its correct position. That means the object number 1 is not at the first position object number 2 will not be at the second position and so.

A derangement refers to a specific type of permutation where no element appears in its original position. For instance, if you have a set of three objects: $A, B, C$, a derangement would be any arrangement where $A$ isn't in the first position, $B$ isn't in the second, and $C$ isn't in the third. Understanding this can be critical when we explore how to calculate the total number of possible derangements for a set of objects.

Imagine you are at a party where you have to wear a name tag with your name on it. If we consider your name as an object, a derangement would mean you switch name tags with others, ensuring you never end up with the tag that has your own name. This creates a more fun and confusing environment.

So, we can divide the set of the derangements of n objects into 2 categories. So, and for these 2 categories, we consider or focus on the element which is there at the first position so we are considering the case where at the first position we have the element a where a could be either a₁, a₂ or a₃… aₙ.

When calculating derangements, we categorize the arrangements based on the element located in the first position. This allows us to address the problem in a more manageable way. The first category involves placing one of the elements in its original position at a different slot, leading to another set of elements that need to be deranged. This structured approach helps in systematically calculating the total number of valid derangements.

Think of an arrangement of books on a shelf. If we need to swap books around so that none of them returns to its original spot, we first choose a book (say, the first one) and move it somewhere else. The choice we make about where to put this book affects how we can rearrange the others, much like how in our method we categorize based on the first element.

Category one, where the element a is occurring at the kth position and remember element a is allowed to be occurring at kth position in a valid derangement, because at the kth position we are not putting a. If that is the case then my problem boils down to the problem of deranging the remaining n - 2 elements.

In this category, once we have placed element $a_1$ in the first position and assigned some element to the kth position, we still need to rearrange all other elements (except the two we've positioned) such that they remain deranged. This means we are left with $n - 2$ elements to consider, since two have been placed and accounted for. The essence of this strategy is essentially reducing the complexity of our problem.

Continuing our book example, let's say we placed book one in the first spot and decided book three can go where book one originally was. Now, we must find new spots for books two, four and any others while ensuring none return to their starting position. This systematic approach allows us to tackle the problem progressively.

The second category of derangements with a occurring at the first position is the following. You do not have the element a allowed at the kth position, that means element a can take any other position of course it cannot take the position a₁.

Here, we account for the fact that the element in the first position cannot occupy the kth position. This restriction means we must consider the possibility of rearranging all remaining elements while ensuring that the first element is still not in its 'home' position. Hence, we consider all other valid arrangements of these elements, which involves deranging $n - 1$ objects rather than just removing the two we’ve fixed.

Imagine if after placing your first book, you state that it can’t go back to its original spot. This opens up the remaining books to fill in their places more creatively, but limits your choices for placement, similar to the constraint introduced in this category.

So, we find out those derangements namely the derangements of n - 1 elements and in that derangement the element a would not be occupying the kth position. You take any such derangement and add the element a at the first position that will now give you a derangement of n objects.

By adding the combination of both categories, we establish a recurrence formula for calculating the total number of derangements. This sum allows us to aggregate the valid arrangements from both scenarios into one total, enabling us to count all possible derangements based on the restrictions identified in positioning elements.

If we consider all the arrangements allowed for our books while ensuring none returns to its starting position, adding together the options from before will settle our overall understanding of how many valid arrangements we could potentially end up with, giving us a complete picture.

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