Find the total number of ways in which $100$ seats can be filled up, provided the $100$th person occupies seat number $100$

Question

There are $100$ people in a queue waiting to enter a hall. The hall has exactly $100$ seats numbered from $1$ to $100$. The first person in the queue enters the hall, chooses any seat and sits there. The $n$th person in the queue where $n$ can be $2,\ldots,100$, enters the hall after $(n-1)$th person is seated. He sits in seat number $n$ if he finds it vacant; otherwise he takes any unoccupied seat. Find the total number of ways in which $100$ seats can be filled up, provided the $100$th person occupies seat number $100$.

My solution goes like this:

According, to the given information, we ignore the $100$ th person and the $100$th seat, as their positions are predetermined. So, we are concerned with only, the persons numbered $1,2,...,99$ and the seats numbered $1,2,3,...,99.$ We consider, $2$ cases as follows:

Case $1$: If person (numbered) $1$ sits in seat (numbered) $1$, then there is only one possible situation under this case, for then person $2$ sits in seat $2$, person $3$ sits in seat $3$ and it continues. So, this case leads us to a particular situation and is possible only in $1$ way. So, this case is possible, only in $1$ way.

Case $2$: This is the case, when person $1$ do not sits in seat $1$. He sits in any other seat, other than seat $1$. Let person $1$, sit in the seat $k_1>1.$ So, $k_1\in\{2,3,...,99\}.$ Now, all the persons numbered $2$ to $k_1-1$, sits in their own places i.e they sit in seats $2$ to $k_1-1.$ The problem occurs, with the $k_1$ th person. When he enters, he finds, the person $1$ in his sit (which is, $k_1$). Now, he has two options:

Option 1: The $k_1$ th person might go to seat number $1$ which is still now, left vacant.

Option 2: The $k_1$ th person can go to any other seat (numbered) $\space k_2\space (\space \text{say}\space)$ such that, $k_2>k_1.$

If $k_1$ th person follows option 1, then the cycle or the process, breaks since, then all the persons numbered from $k_1+1$ to $99$ goes to their original seats,i.e seats numbered from $k_1+1$ to $99,$ and no other problems occur.

Now, if $k_1$ th person follows option 2, then, again, no problem occurs, for the persons numbered from from $k_1+1$ to $k_2-1$ as they sit in the seats $k_1+1$ to $k_2-1$ respectively. But the problem, now occurs again, for the $k_2$ th person. He has again the two options, i.e

Option 1: The $k_2$ th person might go to seat number $1$ if, the $k_1$ th person has not been seated there.

Option 2: The $k_2$ th person can go to any other seat (numbered) $\space k_3\space (\space \text{say}\space)$ such that, $k_3>k_2.$

If the $k_2$ th person follows option 1, then the process breaks and all the rest find their original seats and no further problem occurs. But, now, if the $k_2$ th person follows option $2$ then, no problem occurs for the persons numbered from from $k_2+1$ to $k_3-1$ as they sit in the seats $k_2+1$ to $k_3-1$ respectively. But the problem, now occurs again, for the $k_3$ rd person. The $k_3$ rd person , again, has two options and this process will continue in this case.

So, we observe, we can reduce the problem in case $2$ to : The number of increasing sequences, that we can make, of this form $$2\leq k_1<k_2<k_3<...<k_r<...\leq 99,$$ where $k_i$'s are the persons who faced the problem of finding his/her own seat filled up by someone else. This increasing sequence i.e $$2\leq k_1<k_2<k_3<...<k_r<...\leq 99,$$ can be of any length. If any $k_i's$ (such that $i\neq 2$) goes to seat number $1$ , then, as seen above, the process will hault and so increasing sequence also and we will get the increasing sequence of this form $$2\leq k_1<k_2<k_3<...<k_{i-1}\leq 99.$$ So, the point is, that, depending upon the situation, whether any of the $k_i's$ (such that $i\neq 2$) goes to seat number $1$ or not, we will obtain various increasing sequence, of varying lengths. Now, the total number of ways in which, we can create increasing sequences of the form $$2\leq k_1<k_2<k_3<...<k_r<...\leq 99,$$ is equivalent, to the number of subsets, one can form with set $\{2,3,...,99\}$ and this is equal to $2^{98}$ as ($98$ elements are there and ) each possible such subsets with the.following elements corresponds to an increasing sequence of different lengths( equal to the cardinality of the subset created).

Now, we must note, that among the $2^{98}$ possible ways in, Case $2$, we have, an empty set and this empty set, corresponds to the situation, when we are not building any sequence, or rather, we can state this situation, when we have no such increasing sequence possible. This occurs when, $k_1\ngeq 2$, but this means, $k_1=1,$ so, when we count the empty set, in our calculation in Case $2$, we have included the situation of Case $1$ as well. So, the total number of ways in which, the required arrangement is possible, is $2^{98}$ ways.

Is the above solution valid? Is my procedure correct? If not, where is it going wrong ?...

Answer 1

Disclaimer

This is just an improvement and a little simplification of your own answer.

Answer

Each possible seating corresponds to an increasing sequence $k_{1},…,k_{r}$ such that $2\leq k_{1}\leq …\leq k_{r}\leq 98$.

If the sequence length is non-zero, person $1$ sit in seat $k_{1}$, person $k_{1}$ sit in seat $k_{2}$, and so on until person $k_{r}$ sit in seat $1$. If the sequence length is zero, everyone sit in their own respective seats.

There are $2^{98}$ such sequences

Answer 2

Using the same simplification as in the linked problem... consider the following change to the scenario:

Our first person goes and takes a random seat as normal without looking at their ticket. If a future person $k$ comes and finds the first person in their seat $k$, they kick them out of the seat... thus letting person $k$ sit in their correct seat and forcing our first person to find another seat again from those still open.

The question becomes "How many sequences of selections of seats by our first person exist which do not affect the 100th person?"

The answer to this is that our first person can get popcorned around any subset of the seats labeled 2 through 99 before finally arriving at their intended first seat. There are $2^{98}$ such sequences.