An auditorium has 2 rows of seats in each row . 100 distinguishable people sit in the seats one at a time , Subject to the condition that each person , except for the first person to sit in each row , must sit immediate to the left or rigth of an occupied seat ,and no 2 people can sit in the same seat . In how many ways can this process occur . Answer 2^98×C(100,50)
On
I agree with Lulu that the question is vaguely worded.
However, I believe the key lies in the last sentence," In how many ways can this process occur." In other words, after dividing the people into two groups of $50$ each in $\binom{100}{50}$ ways, in how many orders can the seats numbered $1-50$ in each row be taken, observing the adjacency clause.
For the second part, taking one row of $50$, it is obvious that the last seat taken must either be #$1$ or #$50$. Now let us reverse the process, imagining all $50$ people seated, and take out people one by one. The first out would be #$1$ or #$50$, and supposing you took out #$1$,the next out would be #$2$ or #$50$, and so on, (ie always taking out from either extreme) except that the last person out would have only choice, thus a row can be filled in $2^{49}$ ways.
Thus for $2$ rows and $100$ seats, the answer will be $\dbinom{100}{50}2^{98}$
As already remarked, there is ambiguity regarding the question. I have taken an interpretation that leads to the given answer.
However, if we are to not only consider the order in which rows are filled, but also who fills which row and seat, the answer will be $\dbinom{100}{50}\times2^{98}\times{(50!)}^2$
I will be assuming that the people are indistinguishable even though the "in" is crossed out in the photo. The people will be distinguished by the order in which they sit down.
Step 1
Enumerate all the people from $1$ to $100$. They will sit down in this order.
We can choose who among them sit in the first row in ${100 \choose 50}$ ways.
Step 2
You may assume that number $1$ to $50$ are sitting in the front row.
They decide to form a queue first before sitting down.
Person $1$ starts forming the queue.
Person $2$ can either go immediately in front of person $1$, or immediately behind person $1$.
Person $3$ can then either go immediately in front of person $1$ and $2$, or immediately behind them, and so on.
Person $n$ can either go immediately in front of person $1$, $2$, ..., $n-1$, or immediately behind them.
Finally, person $50$ also chooses where to go, and they have now, in total, made $49$ binary choices when forming this queue. So they end up with one of $2^{49}$ possible ways of forming a queue.
The other $50$ people repeat the process of forming a (separate) queue, something which can also be done in $2^{49}$ ways.
Step 3
Now, magically, everyone realizes that they each had a seat underneat them the entire time.
This gives a total of ${100 \choose 50} \cdot 2^{49} \cdot 2^{49}$ ways of sitting down.