I came across these 2 problem about combinatorics which I originally thought they were basically the same. However, the solution key told me otherwise.
Question 1
Question 2
My answer to Q1 is 0.18 which is correct, I then use the same way to solve Q2, where the solution is 0.2105. I wonder why there is such difference?
In the first case, each customer can choose any of the $20$ cars. The question explicitly says that "at least three units of each car are available", which means that several customers can choose the same (typa-color) combination of car. So the probability is
$$\frac{5\cdot 4}{20}\cdot\frac{4\cdot 3}{20}\cdot\frac{3\cdot 2}{20}=\frac{72}{400}=0.18$$
In the second case, each item is only available once, so three different items are chosen. The probability is
$$\frac{5\cdot 4}{20}\cdot\frac{4\cdot 3}{19}\cdot\frac{3\cdot 2}{18}=\frac{72}{342}\approx 0.2105$$