I'm trying to get a handle on "counting" numbers and outcomes of an event. I am not sure if my math is right, so perhaps someone can show me where I went wrong (unless it's right..).
A hat contains 20 consecutive numbers (1 to 20). If four numbers are drawn at random, how many ways are there for the largest one to be a 16 and the smallest number to be a 5?
The way I solved this was to break it up into four components. Based on the wording of this problem, I am assuming that replacement does not occur.
So, four positions:
____ ____ ____ ____
Position one is reserved for the "16", so there is one possible outcome. Position two is reserved for the "5", so there is one possible outcome there as well. Positions three and four are reserved for the numbers between 5 and 16, with the second position having one less available number since position 3 has "used" that number.
So, my math looks like the following:
n = (1)(1)(10)(9)
n = 90 possible outcomes
Does that sound correct? Is my logic correct?