I'm working on a problem, where 4 people enter a room with 100 seats, and each of these four people has their own chair. However, they seat randomly, so with $\mathbb{P}(\mathrm{own \ chair})= \mathbb{P}(\mathrm{chair\ } i ) =\frac{1}{100}$. Now I want to calculate the probability all four people do not sit at their own seats, multiple occupancy being not allowed. In this post, let $A_i$ denote the event that person $i$ is seated in her/his own chair.
I do this as follows: $\mathbb{P}(\mathrm{no \ one \ at \ their \ own \ chair})=\frac{99}{100}\frac{98}{99}\frac{97}{98}\frac{96}{97}=\frac{96}{100}$.
However, as a hint is given, use the inclusion exclusion principle. So using this, I get: $$\mathbb{P}(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C) = \sum_i \mathbb{P}(A_i^C) - \sum_{i<j}\mathbb{P}(A_i^C \cap A_j^C) +\sum_{i<j<k} \mathbb{P}(A_i^C \cap A_j^C \cap A_k^C) - \mathbb{P}(A_1^C \cup A_2^C \cup A_3^C \cup A_4^C)=4\frac{99}{100}-6\frac{99}{100}\frac{98}{99}+4\frac{99}{100}\frac{98}{99}\frac{97}{98}-(1-\frac{1}{100\cdot99\cdot98\cdot97})=\frac{99!/95!+1}{100!/96!}$$
The outcomes are close, but different. I think that what I did with the inclusion-exclusion principle is wrong. Can anyone say what I did wrong, and how I should use the inclusion-exclusion principle to arrive at the right outcome? Thanks in advance.
EDIT: I thought about the problem, and maybe I should use the Inclusion-Exclusion principe for any intersection of events. Because according to my logic, the probability that no one out of 100 people sits at his own spot is $\frac{1}{100}$, which is certainly wrong, if we compare the probability of a K-matching, for example, or maybe more elementary, that the expected number of correctly seated people is 1, which is strange if $\mathbb{P}(\mathrm{at\ least\ one\ correctly\ seated})=\frac{99}{100}$. That this is impossible again needs K-matching implied probabilities.
Working out this edit, I arrive at the following: $$\mathbb{P}(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C) = \sum_i \mathbb{P}(A_i^C) - \sum_{i<j}\mathbb{P}(A_i^C \cap A_j^C) +\sum_{i<j<k} \mathbb{P}(A_i^C \cap A_j^C \cap A_k^C) - \mathbb{P}(A_1^C \cup A_2^C \cup A_3^C \cup A_4^C)$$ Now calculating the terms on the right hand side gives: $$\begin{eqnarray} \sum_i\mathbb{P}(A_i^C) & = & 4\cdot\frac{99}{100}\\ \sum_{i<j}\mathbb{P}(A_i^C\cap A_j^C) & = & 6(\sum_{i=1}^2\mathbb{P}(A_i^C)-\mathbb{P}(A_1^C\cup A_2^C))\\ & = & 6(2\cdot\frac{99}{100}-(1-\frac{1}{99\cdot100}))=6\cdot\frac{9703}{9900}\\ \sum_{i<j<k} \mathbb{P}(A_i^C \cap A_j^C \cap A_k^C) & = & 4\cdot(-3\sum_{i=1}^3\mathbb{P}(A_i^C)+\sum_{i<j}\mathbb{P}(A_i^C\cap A_j^C)+\mathbb{P}(A_1^c\cup A_2^c\cup A_3^C))\\ & = & 4\cdot(-3\cdot\frac{99}{100}+3\cdot\frac{9703}{9900}+(1-\frac{1}{100\cdot99\cdot98}))\\ & \approx & 4\cdot 0.970302\\ \mathbb{P}(A_1^C \cup A_2^C \cup A_3^C \cup A_4^C) & = & 1- \mathbb{P}(A_1 \cap A_2\cap A_3\cap A_4)=1-\frac{1}{100}\cdot\frac{1}{99}\cdot\frac{1}{98}\cdot\frac{1}{97}\\ \mathbb{P}(A_1^C \cap A_2^C \cap A_3^C \cap A_4^C) & = & 4\cdot\frac{99}{100}-6\cdot\frac{9703}{9900}+4\cdot0.970302-(1-\frac{1}{100\cdot99\cdot98\cdot97})\\ & \approx & 0.9606 \end{eqnarray}$$
I think my answer is correct now. Can anyone verify my calculations?