Example 1-23 There are 9 marked pieces, which are to be placed on a $6 \times 9$ chessboard (Figure 1-7) to simulate the various ways 9 people can exit from 6 exits. Try to discuss how many different schemes there are?
The first cell of each row represents the point facing the exit.
Method 1: Let the 9 pieces $a_{1}, a_{2}, \cdots, a_{9}$ only have labels without any order relationship. $a_{1}$ has 6 choices for the exit; $a_{2}$ has 7 factors to consider, if choosing the same row as $a_{1}$, it can be considered whether to be in front of or behind $a_{1}$. Similarly, $a_{3}$ has 8 possible choices, as $a_{3}$ choosing the same row as $a_{2}$, it must consider whether to be in front of or behind $a_{2}$, and if $a_{1}, a_{3}$ are in the same row, it should also consider whether to be in front of or behind $a_{1}, a_{2}$. By analogy, the number of schemes $N$ is
$$
N=\underbrace{6 \times 7 \times 8 \times 9 \times 11 \times 12 \times 13 \times 14}_{\text {9 items }}=14!/ 5!
$$

Method 2: The markers of the pieces are $a_{1}, a_{2}, \cdots, a_{9}$. Any permutation of them, inserting 5 markers *, can express one

indicating $a_{1}, a_{2}$ exit from the first exit; $a_{3}$ from the second exit; $a_{4}, a_{5}$ from the third exit; $a_{6}$ from the fourth exit; $a_{7}, a_{8}$ from the fifth exit; $a_{9}$ from the sixth exit. For example,

where the fourth exit is empty.
Method 3: As shown in Figure 1-8, 9 pieces $a$ and 5 markers $\bigcirc$. The pieces are marked, while the markers $\bigcirc$ are indistinguishable.
Figure 1-8
[a] and ○ together make 14. The position of ○ is random, choosing 5 from 14 combinations. For example, Figure 1-8 is choosing 3, 6, 8, 10, 12 from $1 \sim 14$. $a_{1} a_{2} \cdots a_{9}$ is a permutation of $1,2,3,4,5,6,7,8,9$. From 14 slots, first choose 5 combinations (for example, 3, 6, 8, 10, 12) to place the markers. The remaining 9 slots, place $a_{i}$ in the empty slots according to the permutation order. Such permutations should be $14!$ kinds, but since ○ are indistinguishable, the arrangement of ○ does not affect, so the same combination corresponds to one exit scheme.
$$
N=9!C(14,5)=\frac{14!}{5!}
$$
The 3 methods lead to the same result.