Two friends board an airliner just before departure time. There are only 9 seats left, 4 of which are aisle seats. How many ways can the 2 people arrange themselves in available seats so that at least one of them sits on the aisle?
The 2 people can arrange ihemselves in ways.

Step 1 ：Consider two cases: 1. Both friends sit in aisle seats. 2. One friend sits in an aisle seat and the other does not.

Step 2 ：For the first case, we have 4 aisle seats and we need to choose 2 of them for the friends. This is a combination problem, and we can use the combination formula $C(n,k)=\frac{n!}{k!(n-k)!}$, where n is the total number of items, and k is the number of items to choose. The number of ways for this case is $C(4,2)=6$.

Step 3 ：For the second case, one friend sits in an aisle seat and the other does not. We have 4 aisle seats and we need to choose 1 of them for the friend who sits in the aisle. The other friend can choose any of the remaining 8 seats (5 non-aisle seats and 3 aisle seats). This is also a combination problem. The number of ways for this case is $C(4,1)\ast C(8,1)=32$.

Step 4 ：The total number of ways is the sum of the number of ways in these two cases, which is $6+32=38$.

Step 5 ：Final Answer: The total number of ways the two friends can arrange themselves in available seats so that at least one of them sits on the aisle is $\boxed{{\displaystyle 38}}$.