Question: An urn contain 9 balls three red four blue and two are green. Three balls are drawn at random without replacement from the urn. What is the probability that all three balls different colour.
Solution
The way of solving it which I saw on youtube was : $$\tag{1} {3 \choose 1} \times {4 \choose 1} \times {2 \choose 1}/{9 \choose 3},$$
which gives the answer $2/7$.
How I like to think about this is, since it is without replacement so $$\tag{2}(\text{prob. of red balls}) (\text{prob. of blue balls}) (\text{prob. of green ball}) = \frac 39 \frac 48 \frac 27 \times 6 ;$$
multiplying with six as there can be 6 possible arrangements of three different balls.
Where I am stuck : I am unable to picture what's happening in 1st formula.
If I see it like probability = (favorable outcom)/(total outcomes)
So does ${ 9\choose 3}$ telling me the sample space?
If yes, how? ${9 \choose 3}$ is selection it does care about order, like 'abc' and 'bca' are same for combination, so how can it be sample space.
If I am using any wrong logic please help.
The important point to note is that the count for the numerator and denominator must be done in a consistent manner.
You may best understand it by combinations vs permutations
When you are using the formula $\large\frac{\binom31\binom21\binom41}{\binom93}$,
you have used combinations in both numerator and denominator, so there is consistency.
But when you simply write $\large \frac39\frac48\frac27$, the numerator is essentially a combination $\binom31\binom41\binom21$, while the denominator is a permutation $9\cdot8\cdot7 = ^9P_3$
so we multiply the numerator by $3!$ to also make the numerator a permutation