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Can the bijection for the Lost Boarding Pass Probability Problem, be formulated or pictured?

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

2. 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](https://math.stackexchange.com/questions/5595/taking-seats-on-a-plane/592225#592225)

>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.](https://old.reddit.com/r/math/comments/d8qp0/a_combinatorics_problem_that_i_was_never_able_to/c0yei3a/)

> 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.