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The two proofs below moot "bijection", but I don't see a formula. 1. Does the bijection have an explicit formula?

  1. Can this bijection be pictorialized? My 16 y.o. kid doesn't understand the abstract, Daedalian phrasing below.

hunter's answer dated Dec. 4 2013

Claim 3: There is a bijection between the set of admissible seatings in which the final passenger gets his seat and the set where he doesn't.

Proof: Suppose for an admissible seating $S$ that passenger $n$ is the first to choose one of {first passenger's seat, last passenger's seat}. By claim $2$, there is a unique admissible seating $T$ which agrees with $S$ except that passenger $n$ and the final passenger make the opposite decision ($T$ matches $S$ until passenger $n$ sits, then by Claim 2, $T$ must continue to match $S$ until the final passenger).

azjps comments on A combinatorics problem that I was never able to solve.

Here's a bijective proof: let all $n$ individuals find a seat according to the rule. For each sequence in which the $n$th person sits in the seat assigned to the $k$th person, consider the corresponding sequence in which the $n$th and $k$th individuals are swapped (so the last person is in the right seat). Both sequences have equal probability and it is not difficult to verify that this correspondence is a bijection, so the answer is 1/2.

  • Using the claim that the last passenger always sits at the first or last seat, you can make the bijection between $(n, i_1, i_2, ..., i_{n-1})$ and $(i_{n-1}, i_1, i_2, ..., n)$. These two have the same probability (it just depends in the choice of the person with number $i_{n-1}$), one has $n$ in their seat and another doesn't. You can also write or draw some examples for the case where there are only $4$ people so it's maybe clearer. – AnilCh Jul 27 '21 at 07:22
  • For the first, the bijection is just that given an admissable assignment where the last seat is assigned one way, there exists a unique admissable assignment agreeing except with two seats flipped (the $n$th and the one for the last passenger). You can't really visualize this since there is, a priori, no geometric structure. Instead of visualizing bijections, use the definition: a bijection $A \cong B$ is a pairing such that every element $a \in A$ is paired with a unique element $b \in B$ and such that $b$ is uniquely paired with $A$. – Brevan Ellefsen Jul 27 '21 at 07:24
  • In this case, $S$ and $T$ are the elements of two sets, namely the admissable assignments where the last passenger sits in her own seat, and the admissable assignments where she sits in the first passenger's seat. Since two seats are permuted, we also see that if we had chosen $T$ first then we would get $S$ after permuting the seats, so that $S$ and $T$ are paired under the bijection we are building. Since this holds for all possible $S$ and $T$, and any choices $S, S'$ with $S \neq S'$ will yield distinct paired elements $T,T'$ (they will differ in the non-permuted seats), we are done. – Brevan Ellefsen Jul 27 '21 at 07:26

1 Answers1

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I don't know whether this counts as picturing the bijection, but let's write out all the admissible seatings for a four-passenger plane. We will represent each admissible seating by the cycle of displaced passengers, numbered according to their position in line, with passenger $1$ counting as displaced whether they end up sitting in their own seat or not. The cycle $(1,a,b,c,d)$, for example, will mean $1$ sits in $a$'s seat, $a$ sits in $b$'s seat, $b$ sits in $c$'s seat, $c$ sits in $d$'s seat, and $d$ sits in $1$'s seat. It is necessary that $1<a<b<c<d$. $$ \begin{array}{r|r||r|r||r} \text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\ (1) & 1234 & (1,4) & 4231 & \frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{4}\\ (1,2) & 2134 & (1,2,4) & 4132 & \frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{12}\\ (1,3) & 3214 & (1,3,4) & 4213 & \frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{8}\\ (1,2,3) & 3124 & (1,2,3,4) & 4123 & \frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{24}\\ \hline & & & & \frac{1}{2} \end{array} $$ In the first row of this table, the cycle $(1)$ represents the situation where passenger $1$ sits in her own seat, while the cycle $(1,4)$ represents the situation where passenger $1$ sits in passenger $4$'s seat and passenger $4$ sits in passenger $1$'s seat. Each of these situations occurs with probability $\frac{1}{4}$. Similarly in row $2$, the cycle $(1,2)$ represents the situation where $1$ sits in $2$'s seat and $2$ sits in $1$'s seat, while the cycle $(1,2,4)$ represents the situation where $1$ sits in $2$'s seat, $2$ sits in $4$'s seat, and $4$ sits in $1$'s seat. Each of these situations occurs with probability $\frac{1}{12}$. The remaining two rows can be understood similarly. Observe that the total probability of the situations represented by the first column and the total probability of the situations represented by the third column both equal $\frac{1}{2}$. In column $1$, passenger $4$ sits in his own seat; in column $3$, passenger $4$ sits in passenger $1$'s seat.

The admissible seatings for a five-passenger plane are below. $$ \begin{array}{r|r||r|r||r} \text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\ (1) & 12345 & (1,5) & 52341 & \frac{1}{5}\cdot1\cdot1\cdot1\cdot1=\frac{1}{5}\\ (1,2) & 21345 & (1,2,5) & 51342 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{20}\\ (1,3) & 32145 & (1,3,5) & 52143 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{15}\\ (1,4) & 42315 & (1,4,5) & 52314 & \frac{1}{5}\cdot1\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{10}\\ (1,2,3) & 31245 & (1,2,3,5) & 51243 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{60}\\ (1,2,4) & 41325 & (1,2,4,5) & 51324 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{40}\\ (1,3,4) & 42135 & (1,3,4,5) & 52134 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{30}\\ (1,2,3,4) & 41235 & (1,2,3,4,5) & 51234 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{120}\\ \hline & & & & \frac{1}{2} \end{array} $$

Will Orrick
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