When do we need combination factor? Say I want to draw 4 balls from a big ball pool, with 3 kinds of colors:
red 50%, white 30%, black 20%.
Now, I draw 1 ball of each time for 4 times, each time with replacement(or the pool is big enough that you don't need to consider this effect) 
The observation: red,red,black,white
My friends and I have two kinds of calculation for the probability of this observation:
Number 1: Consider this as drawing 4 balls from a multinomial distribution:
P(observation)=$4!/(2!1!1!)*(0.5)^2*(0.3)^1*(0.2)^1$
Number 2: Some of us think that the observation is what it is, which is fixed, 
so the combination factor 4!/(2!1!1!) is not needed, so that:
P(observation)=$(0.5)^2*(0.3)^1*(0.2)^1$
This puzzles me a lot, when do we need to consider the combination factor and when we don't?
 A: It's a matter of whether order is important.
Without the combinations factor you are only calculating the probability  of one particular sequence of drawing the balls ( e.g. first 2 balls red, then one white followed by one black ) . To include all possible sequences you need the factor.
A: Here are different ways of looking at the question:


*

*The probability you have seen that observation:

*

*is $1$, since you saw it


*The likelihood of the observation:

*

*is proportional to $p_\text{red}^2p_\text{black}p_\text{white}$ and this proportionality means it does not matter whether you include the factor of $12$ or not


*The significance of the observation:

*

*involves adding up the probabilities of seeing results as extreme or more extreme than that observation, and will also include other possible patterns you did not observe; the $12$ different orders are each as extreme as the particular one you saw


*The probability of the next observation being the same:

*

*is $0.18$ if order does not matter in deciding sameness 

*is $0.015$ if order matters in deciding sameness
