Twelve students from three CCAs organised an overseas learning trip to Shanghai. It comprises five students from Tennis, four students from Bowling and three students from Softball. The available seats for the flight for them to choose from are shown on the diagram as follows:

1. Assign the particular student to the fixed seat

The particular student is assigned to the fixed seat nearest to the emergency exit (Row 55 Seat I). There is $1$ way to do this, leaving $11$ students and $11$ available seats.

2. Arrange the remaining students in the remaining seats

The remaining $11$ students can be arranged in the remaining $11$ available seats in $11!$ ways.
$\text{Total number of ways} = 1 \times 11! = 39\,916\,800$

1. Identify the available non-aisle seats

Aisle seats are columns C, D, F, G. The non‑aisle seats lie in columns E, H, I. The available (un‑crossed) non‑aisle seats are:

$\text{Row }55\,E,\ \text{Row }57\,E,\ \text{Row }55\,H,\ \text{Row }56\,H,\ \text{Row }55\,I,\ \text{Row }56\,I$

This gives $6$ non‑aisle seats available for the two restricted students.

2. Seat the two restricted students first

The two particular students must take $2$ of the $6$ non‑aisle seats, and order matters (different students in different seats):

$^6P_2 = 6 \times 5 = 30 \text{ ways.}$

3. Seat the remaining students and combine

After seating the two students, $10$ seats remain for the other $10$ students:

$10! = 3\,628\,800 \text{ ways.}$

By the multiplication principle:

$^6P_2 \times 10! = 30 \times 3\,628\,800 = 108\,864\,000.$