If we employ the counting method applied in the question

tough-p-n-c-92675.html then

Considering three different samples and contestant not to taste all three samples there would be following two conditions

1. Contestant tastes 3 cups of on sample and 1 cup of another

i.e. AAAB

The number of ways this can happen is

3C2*4!/3! = 12

3C2 = Ways to select any 2 samples out of the 3 choices

4!/3! = # of ways AAAB can be rearranged

or

2. Contestant tastes 2 cups of a sample and 2 of another

i.e. AABB

The number of ways this could happen is

3C2*4!/(2!*2!) = 18

3C2 = Ways to select any 2 samples out of the 3 choices

4!/(2!*2!) = # of ways AABB can be rearranged

Hence total number of ways by which contestant can only taste any two samples is 18 + 12 = 30

Total number of ways to select 4 cups from 9 = 9C4 = 126

hence the probability that contestant can only taste any two samples = 30/126

= 5/21

Is this approach not correct? Where am I missing ?

I have come up with another approach which gives a different result again.

Assuming the that the contestant does not taste the 3rd sample there are 2 scenarios possible

Counting one at a time.

Hence the total number of ways in which the contestant cannot taste any cup from third sample = 9 +27 = 36

Even after hours and hours of going through various material on Probablity and combinations I am struggling to identify a uniform approach that can be applied across problems. How would one know which approach to apply for which problem during the exams. Is there any way we can determine that?