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A little background to my question:-

After starting to read the inclusion and exclusion principle I encountered the topic of derangement of objects and as I was not able to come up with a proof for the formula for derangement of $n$ objects(the book posed the proof of derangement as an exercise), which is

$$\mathcal{D}(n)=n!\sum_{k=0}^n\frac{(-1)^k}{k!}$$

Then after searching for some good hints here I stumbled upon robjohn's fabulous answer here. After going through all the methods that were mentioned there and after irritating robjohn with a lot of questions I ended up with a final question regarding the third method, which has an approach of PIE. After thinking about my question a lot(which you will see in the next section) I thought it was a conceptual doubt regarding PIE and did not belong to the comment section of robjohn's answer, hence I am creating a new post.

My question regarding Derangement:-

After discussing all the steps that robjohn had taken in deriving the formula for derangement in his third method in the answer I tried to recreate the proof on my own with some slight variations to come up with explanations so as to fully understand the concept of derangement. So, here is my approach.

Consider the set $A_i$, which gives the different permutations of the $n$ objects keeping the $i^\text{th}$ object fixed.

Then, $\left|A_i\right|$ counts all the different permutations in which only the $i^\text{th}$ object remains fixed. So $\left|A_i\right|=(n-1)!(\text{ways to arrange the remaining $(n-1)$ objects})$

Similarly, $\left|A_i\cap A_j\right|$ counts all the different permutations in which only the $i^\text{th}$ and $j^\text{th}$ objects remains fixed. Hence, $\left|A_i\cap A_j\right|=(n-2)!$

So, in general $$\left|\bigcap_{i=r}^k{A_i}\right|=(n-(k-r+1))!\qquad\qquad\qquad (k,r\lt n)$$

Now, comes the part where I am getting confused and which will supplement to the incomplete heading of the post.

To count the number of derangement of the $n$ objects why is it that we have to consider the cardinality of the set $\displaystyle{\bigcap_{i=1}^n A_i^c}$ and not $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$.

With the above question what I want to ask is what is the difference between $\displaystyle{\bigcap_{i=1}^n A_i^c}$ and $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$.

What I understand from $\displaystyle{\bigcap_{i=1}^n A_i^c}$ is that that it accounts for all those permutations in which neither the $1^\text{st}$ nor the $2^\text{nd}$ nor the $3^\text{rd}$ $\ldots$ nor the $n^\text{th}$ object occupy their respective position. In short none of the objects occupy their respective positions, precisely what derangement is.

But then, $\displaystyle\left(\bigcap_{i=1}^n A_i\right)$ accounts for all those permutations in which all the object occupy their respective positions, so $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$ accounts for all those permutations in which none of the objects occupy their respective positions.

I know that the way I interpret $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$ is wrong, since $\displaystyle\left|\bigcap_{i=1}^n A_i\right|=1$. But, what is wrong in my interpretation that I couldn't find out

user350331
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1 Answers1

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Then, $\left|A_i\right|$ counts all the different permutations in which only the $i^\text{th}$ object remains fixed.

No, $|A_i|$ is the number of permutations in which the $i$-th and possibly other elements are fixed.

Similarly, $\left|A_i\cap A_j\right|$ counts all the different permutations in which only the $i^\text{th}$ and $j^\text{th}$ objects remains fixed.

$|A_i\cap A_j|$ is the number of permutations in which at least the $i$-th and $j$-th elements are fixed, and possibly others as well.

To count the number of derangement of the $n$ objects why is it that we have to consider the cardinality of the set $\displaystyle{\bigcap_{i=1}^n A_i^c}$ and not $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$.

For each $i$, $A_i^c$ is the set of permutations that do not fix the $i$-th element. The intersection of these sets is therefore the set of permutations that do not fix the $i$-th element for any $i$, i.e., the set of derangements. The set $\bigcap_iA_i$, on the other hand, is the set of permutations that fix every element; it contains exactly one permutation, the identity permutation, and its complement, $\left(\bigcap_iA_i\right)^c$, contains the other $n!-1$ permutations. Your mistake is here:

But then, $\displaystyle\left(\bigcap_{i=1}^n A_i\right)$ accounts for all those permutations in which all the object occupy their respective positions, so $\displaystyle\left(\bigcap_{i=1}^n A_i\right)^c$ accounts for all those permutations in which none of the objects occupy their respective positions.

The negation of all objects are fixed is not no objects are fixed: it’s at least one object is not fixed.

Brian M. Scott
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