Final answer:
There are 126 ways David can pick four numbers from the first twelve positive integers so that no two of the numbers are consecutive, using the combination formula.
Step-by-step explanation:
The student is asking how many ways David can pick four numbers from the first twelve positive integers so that no two of the numbers are consecutive. To solve this, we must consider that after picking any number, the next possible number David can choose needs to be at least two numbers away to avoid consecutiveness.
Let's denote the 12 positive integers as spaces where David can place his four picks: _ 1 _ 2 _ 3 _ 4 _ 5 _ 6 _ 7 _ 8 _ 9 _ 10 _ 11 _ 12 _. We can see that there are 13 slots where David can place a pick (indicated by underscores), but because he needs to pick four while avoiding two consecutive numbers, we effectively need to place three additional slots after each pick to avoid consecutiveness. This turns our problem into finding the number of ways to place four items into nine slots.
Using the combination formula C(n, k) = n! / (k!(n - k)!), where n is the number of total items and k is the number of positions to choose, we have C(9, 4) = 9! / (4!(9 - 4)!) = 126 as there are 126 ways to pick the four numbers such that none are consecutive.