Total Sample Space Count - Permutation or Combination? Consider the scenario that you have 50 balls and let an experiment be that 5 balls are randomly selected. How can I calculate the total number of sample points (the elements in the sample space)?
Should I use 50C5 (50 choose 5) or should I do 50P5 (50 permute 5)? I've mostly seen "combination" being used but I can't understand the essence of it.
For the first ball, we have 50 options, and for the next we have 49 ways, for the next we have 48 ways to choose a ball and so on. From the fundamental rule of counting, it should be 50x49x48x47x46 (which is 50P5).
 A: Give it an analogy (or a real life example): You're a coach of an NBA team with 50 players. You need to choose 5 players to play tonight. In how many ways can you form a team of 5?
If you choose: A,B,C,D & E to play it's identical to choosing E,D,C,B & A. Point is: it doesn't matter which one comes first or last - you just need 5 players. 
For this - you use Combination! - mind the 5! that eliminates all same combo.
$$\binom{50}{5} = \frac{50!}{45!5!} $$ 
However, in permutation the order is important. Back to the NBA: in how many ways can you choose 5 players from a poll of 50 players where the first player is considered MVP the second player is vice-MVP and so forth? Answer: 50x49x48x47x46
A: For the sake of demonstrating N. F.'s point in the comments, let's suppose the balls are labelled 1 through 50 and a "success" is choosing balls 1 through 5.
If you choose the balls one at a time (so that they can be distinguished by their order), then the probability of a success is
$$
\frac{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{50 \cdot 49 \cdot 48 \cdot 47 \cdot 46} = \frac{1}{2118760}.
$$
If you choose the balls five at a time (so that they cannot be distinguished by their order), then the probability of a success is
$$
\frac{5 C 5}{50 C 5} = \frac{1}{2118760}.
$$
As long as you are consistent in your application of order in both the numerator and the denominator, you get the same probability. (This is actually quite a useful observation, as many probability calculations are much simpler if you suppose that order matters.)
