来自三个 CCA（课外活动组织）的 12 名学生组织了一次去上海的海外学习之旅。其中包含 5 名网球社的学生、4 名保龄球社的学生和 3 名垒球社的学生。他们可选择的航班空余座位如下图所示：

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:

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