Explain to a 15 y.o. Byron Schmuland's answer that uses Summation and Product notations to solve the Crazy Lady Airplane Seat probability problem? Byron Schmuland's answer1 is too abstruse for a 15 y.o. student who needs details! Please see my enumerated questions below.

Let's find the chance that any customer ends up in the wrong seat.
For $2\leq k\leq n$, customer $k$ will get bumped when he finds his seat
occupied by someone with a smaller number, who was also bumped
by someone with a smaller number, and so on back to customer $1$.
This process can be summarized by the diagram
$$1\longrightarrow j_1\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k.$$
Here $j_1<j_2<\cdots <j_m$ is any (possibly empty) increasing sequence of integers strictly
between $1$ and $k$.
The probability of this sequence of events is
$${1\over n}\times{1\over(n+1)-j_1}\times {1\over(n+1)-j_2}\times\cdots\times{1\over(n+1)-j_m}.$$


*

*What exactly does $1/n$, the first term, signify?


*Where do each of the ${1\over(n+1)-j_m}$ terms hail from and signify? How would you divine or forebode to formulate this term?

Thus, the probability that customer $k$ gets bumped is
$$p(k)={1\over n}{\color{red}{\sum}}\prod_{\ell=1}^m  {1\over(n+1)-j_\ell}$$ where the sum is over all sets of $j$ values $1<j_1<j_2<\cdots <j_m<k$. That is,



*Where did the $\color{red}{\sum\limits_{1<j_1<j_2<\cdots <j_m<k}}$ stem from? It appears to come out of left field!


\begin{eqnarray*}
p(k)&=&{1\over n}\color{limegreen}{\sum_{J\subseteq\{2,\dots,k-1\}}}\ \, \prod_{j\in J}{1\over (n+1)-j}\cr
    &=&{1\over n}\ \, \prod_{j=2}^{k-1} \left(1+{1\over (n+1)-j}\right) \cr 
    &=&\color{sienna}{{1\over n}\ \,\prod_{j=2}^{k-1} {(n+2)-j\over (n+1)-j}}\cr
    &=&\color{sienna}{1\over n+2-k}.
\end{eqnarray*}



*How does $\color{red}{\sum\limits_{1<j_1<j_2<\cdots <j_m<k}} \equiv \color{limegreen}{\sum\limits_{J\subseteq\{2,\dots,k-1\}}}$?


*How does $\color{sienna}{{1\over n}\ \,\prod\limits_{j=2}^{k-1} {(n+2)-j\over (n+1)-j}={1\over n+2-k}}$?

In the case $k=n$, we get $p(n)=1/2$ as in the other solutions. Maybe there is an intuitive explanation of the general formula; I couldn't think of one.




Added reference: Finding your seat versus tossing a coin by Yared Nigussie, American Mathematical Monthly 121, June-July 2014,  545-546.
1I currently see no answer on this page by someone called "Byron Schmuland", but this other question refers to the answerer as Byron Schmuland, and there's merely one deleted user. Then I deduced that user940 (the deleted user) was University of Alberta Prof. Byron Schmuland PhD Carleton University 1987.
 A: So basically the original author meant to find the probability that:

*

*First passenger randomly choose passenger $j_{1}$’s seat. Since there are $n$ seats to choose, the probability is $\frac{1}{n}$. This is the answer to your first question.

*Passenger $j_{i}$ who got kicked from his original seat randomly choose passenger $j_{i+1}$’s seat. By the time passenger $j_{i}$ need to sit $j_{i}-1$ passengers have sit, so there are  $n+1-j_{i}$ to choose. The probability of choosing a particular seat is then $\frac{1}{n+1-j_{i}}$. This is the answer to your second question.

Recall that $\frac{1}{n}\prod_{i=1}^{m}{\frac{1}{n+1-j_{i}}}$ is only the probability of a single possibilities i.e. the probability that the first passenger along with passenger $j_{i}\phantom{x}\forall\phantom{x}1\leq i\leq m$ sit in the wrong seat.So you need to add the probability of all possibilities. This is the anwer to your third question.
Since $1<j_{1}<…<j_{m}<k$ then $\{j_{1},…,j_{m}\}$ is a subset of $\{2,3,…,k-1\}$. This is the answer to your fourth question.
Hint for your fifth question: telescoping multiplication.
