Answer:
the airline can offer 20 different flight paths under the given conditions.
Explanation:
If the airline has 6 airports and the plane lands three times without staying in the same place twice, this is essentially a permutation problem. You want to find the number of ways to arrange 3 distinct airports out of 6.
This can be calculated using the formula for permutations of "n" items taken "r" at a time:
nPr = n! / (n - r)!
Where "n" is the total number of items (airports in this case) and "r" is the number of items to be arranged (3 landings in this case), and "!" denotes factorial.
So, in your case, the calculation would be:
6P3 = 6! / (6 - 3)!
= (6 * 5 * 4 * 3 * 2 * 1) / (3 * 2 * 1)
= 120 / 6
= 20
So, the airline can offer 20 different flight paths under the given conditions.