1. The first car has 6 choices, the second car has 7 choices, because when the second car chooses the same lane as the first car, there are two cases: before or after the first car; the third car has 8 choices... and so on, we get
$$
N=6 \times 7 \times 8 \times 9 \times 10 \times 11 \times 12 \times 13 \times 14=\frac{14!}{5!} .
$$

Alternative method: $a_{1} \cdot a_{2} a_{3} \cdot a_{4} a_{5} \cdot a_{6} \cdot a_{7} a_{8} \cdot a_{9}$
There are 14 elements, 9 $\square$s, and 5 -s. Determining 5 -s in 14 positions gives $\mathrm{C}_{14}^{5}$ ways. Once the -s are determined, fill in the cars in the order $a_{1} a_{2} \cdots a_{9}$ into the $\square$s, which results in one high-speed entry scheme. Therefore, $N=C_{14}^{5} \cdot 9!=\frac{14!}{5!}$.
1. At a highway entrance, there are 6 lanes for vehicles to enter. Now, 9 cars are about to enter the highway. How many ways are there for them to enter?