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Given the set $\Omega$ containing the permutations of the numbers $\{1,2,...,N\}$, we consider the uniform distribution $\mathbb{P}$, so that the probability of an elementary event, i.e. the probability of drawing an arbitrary permutation will be $\mathbb{P}({\sigma}) = 1/N!.$ A fixed point of a permutation $\sigma$ is a number $i\in \{ 1,2,...,N \}, \sigma(i) = i.$ I would like to find the probability to draw at random a permutation containing $k, 0\leq k \leq N,$ fixed points by using the inclusion-exclusion principle, $$\mathbb{P}(\cup_{i=1}^n S_i) = \sum_{k=1}^n (-1)^{k-1} \sum_{1\leq i_1 < \dots <i_k\leq n} \mathbb{P} (S_{i_1} \cap \dots \cap S_{i_k}).$$ The number of permutations with $k$ fixed points is $C_k^N = \frac{N!}{k!(N-k)!},$ so that the probability of getting at random a permutation with $k$ fixed points will be $\frac{C_k^N}{N!} = \frac{1}{k!(N-k)!}.$ Is this correct ? How can I find this by using the inclusion-exclusion principle ? I thought about the following: one needs to divide the set of permutations containing $k$ fixed points into not necessarily disjoint subsets. For example one way would be to consider the permutations (with $k$ fixed points) which contain $1$ as a fixed point all over to permutations which contain $N$ as a fixed point. I do not know how to go from here. Do you have any suggestion or a solution proposal ? Thanks.

user996159
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    First, you should get familiar with understanding derangements and how inclusion-exclusion is used for that. With this as a foundation, and denoting the number of derangements of $m$ elements as $!m$, the answer to your question of the probability of $k$ fixed points when looking at a random permutation of $N$ elements will be $\dfrac{\binom{N}{k}\cdot !(N-k)}{N!}$, seen by choosing the fixed points and then choosing a derangement of the rest. – JMoravitz Sep 20 '23 at 12:43
  • Thanks. I do not understand why my result deviates from this one. But I am unfamiliar with the subject of derangements. – user996159 Sep 20 '23 at 12:57
  • As to your attempt, you say "The number of permutations with $k$ fixed points is $\binom{N}{k}$" This is inaccurate for a number of reasons. That is merely the number of ways of selecting $k$ points to fix in place, but you still need to consider the number of ways in which the remaining $N-k$ points can be arranged. You might be inclined to say that you have $(N-k)!$ ways to arrange them, but in some of those you will have additional fixed points. – JMoravitz Sep 20 '23 at 13:10
  • The punchline (as explained in the linked post) is for those $N-k$ remaining elements, consider "The first is not assigned to itself and the second is not assigned to itself and ..." as the complement of "The first is assigned to itself or the second is assigned to itself or ..." and run inclusion-exclusion on that. – JMoravitz Sep 20 '23 at 13:12
  • Thanks. I understand the mistake. But I am still not able to use the inclusion-exclusion principle to find the same result. – user996159 Sep 20 '23 at 15:01
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    Without more information, I can not tell you what you are doing wrong or missing. I can only emphasize that the linked post (and the many other posts linked to that post) already contains many details on how to calculate derangements by many methods, including inclusion exclusion, and that my writing as $!(N-k)$ is shorthand for the final result rather than the expression that would have come about from inclusion-exclusion, having been $\sum\limits_{i=0}^{N-k} (-1)^i\binom{N-k}{i}(N-k-i)!$ or similar (in case I made a typo). The shorthand is far easier to read and write. – JMoravitz Sep 20 '23 at 15:11
  • What is the meaning of this phrase from the link, " Then the number of permutations in $k$ of the $S_i$ would be the number of permutations that fix $k$ items."? Also, when I say inclusion-exclusion principle, I mean the expression I have posted in my question. Following the link, I can not see how such a principle is used there. I will appreciate your help. Thanks. – user996159 Sep 20 '23 at 17:13
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    The sentence right before that one says $S_i$ is how they are labeling the set of permutations who fixed element $i$. If a permutation is an element of exactly $5$ of these $S_i$ then it must be a permutation who fixed exactly five elements. Similarly if we use an arbitrary number $k$. "When I say inclusion-exclusion I mean the expression..." It appears in the very next line... just where we don't use $\sum\limits_{1\leq i_1<i_2<\dots<i_k\leq n}$ since its not necessary since all ways of choosing $k$ indices result in the same count. We got to replace that ugly sum with $\binom{n}{k}$ – JMoravitz Sep 20 '23 at 17:37
  • Where does the factor $(-1)^{k-1}$ come from in the link ? May I assume that $S_i$ in the link contains a single fixed element, the element i, (this is not stated clearly in the link) ? – user996159 Sep 20 '23 at 18:56
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    The powers of $-1$ come from using inclusion-exclusion. The $S_i$ will certainly fix element $i$ but may fix more than just that. – JMoravitz Sep 20 '23 at 21:26

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