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The question is as follows:

Harvard Law School courses often have assigned seating to facilitate the “Socratic method.” Suppose that there are $100$ first year Harvard Law students, and each takes two courses: Torts and Contracts. Both are held in the same lecture hall (which has $100$ seats), and the seating is uniformly random and independent for the two courses. Find the probability that no one has the same seat for both courses (exactly; leave the answer as a sum).

Here is my attempt:

$$P(no\; same\; seat) = 1 - P(all\; same\; seats)=1-(\frac{1}{100}+\frac{1}{99}+...+\frac{1}{1})=1-\sum_{i=1}^{100}{\frac{1}{i}}$$

So my thinking goes like this. The $1st$ student can choose any seat in the first class, thus, to choose the same seat in the second class would be $\frac{1}{100}$. The $2nd$ student can choose any of the $99$ remaining seats for the first class, thus, to choose the same seat in the second class it would be $\frac{1}{99}$.

However, upon looking at the solution, it seems like my answer is completely wrong. Does anyone know why my method is wrong?

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    $\begingroup$ To begin, $Pr(all~same~seats)$ would be $\frac{1}{100}\times \frac{1}{99}\times\cdots$ not $\frac{1}{100}+\frac{1}{99}+\cdots$. Next, the opposite event of "No same seats" is not "All same seats." The correct opposite event is "At least one same seat." $\endgroup$
    – JMoravitz
    May 26, 2016 at 19:19
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    $\begingroup$ Letting each student be labeled by the seat they picked in the first class, the question is essentially asking what the probability is of a Derangement occurring when considering the seating for the second class. $\endgroup$
    – JMoravitz
    May 26, 2016 at 19:20
  • $\begingroup$ Also, the expression for the probability in the attempt is $< 0$, a sure sign that something has gone awry. $\endgroup$ May 26, 2016 at 19:24
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    $\begingroup$ This is somewhat related: math.stackexchange.com/questions/1799767/… $\endgroup$ May 26, 2016 at 19:24

1 Answer 1

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The problem is known as derangements, usually presented as passengers with tickets randomly assigned seats on a plane or guests receiving the wrong hats. Solution to it is an inclusion-exclusion principle/theorem. Obviously the total number of seatings is $100!$. How many do we need to exclude to get all 'wrong' ones? Assume at least 1 person has a correct seat. You have $\binom{100}{1}$ such allocations. For each there are $99!$ seatings for other students. After you've subtracted this from $100!$ you need to add back cases where at least two students, $\binom{100}{2} \times 98!$ and so on. Can you take it from here?

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