Consider the case of choosing 3 marbles from a bag of 10 containing
3 red, 5 green and 2 blue marbles.
Find the probability of choosing exactly 2 red marbles and 1 green marble without replacement.
The standard (or rather, the memorized) answer to this type of question is
$$\frac{^3C_2~^5C_1}{^{10}C_3}$$
But I've never been able to fully understand the logic behind this formulation- my interpretation of the denominator is that-
"let's pretend that all the marbles are distinct by labeling each with a number 1-10, to find out the number of ways in which these 10 "distinct" marbles (or numbers) can be arranged into three places, we compute $^{10}P_3$ since the order in which the marbles (numbers) are picked has no specific meaning (i.e. RRG is treated to be the same as RGR- *is this inference even correct? (by intuition the order does matter, since being the first/second/last one to be chosen does have some specific meaning?)), we write $\frac{^{10}P_3}{^3P_3}={^{10}C_3}$"
*But where in the denominator takes into account that "in fact, in these 10 marbles, we know that there are 3 identical red marbles, 5 identical green…"?
Similarly for the numerator:
"Pretend the 3 red marbles (5 green marbles) are all distinct by labeling them each with a number 1-3 (1-5), $^3C_2$ ($^5C_1$) gives the number of ways in which the 3 (5) distinct numbers can be arranged into 2 (1) places where the order does not matter"
*What in the numerator has fixed the meaning to give "actually, the three (five) red (green) marbles are all identical even though they're all labelled with a number."?
*Also is there a quick way of knowing when the order matters and when it doesn't?
(My understanding on when the order matters is limited to when we're talking about setting up a PIN, the distribution of gold/silver/bronze medals, and it doesn't matter when all the prizes are the same- but it seems quite insufficient to help me understand what makes the order matter).
Best Answer
We don't have to pretend the marbles are distinct, they are in fact different marbles. It is just that sometimes we don't care to distinguish between them save by colour, or not even by that.
How many ways can we select 2 from 3 red marbles: Let's arrange the three marbles and select the first two: $$\require{enclose}\def\C#1{\color{red}{\enclose{circle}{\color{black}{#1}}}}\begin{array}{l:l}\C 1\C 2\quad \C 3 & \C 2\C 1\quad \C 3 \\ \hdashline \C 1\C 3\quad \C 2& \C 3\C 1\quad \C 2\\ \hdashline\C 2\C 3\quad \C 1& \C 3\C 2\quad \C 1\end{array}$$
So there are $6$ ways... but wait, in some of these the selected marbles are the same but in different orders, but we don't care about the order in which they are selected, only that they are. So there are $3!/2!1!$ ways to select two from three red marbles.
A heap of $n$ elements can be arranged in $n!$ ways, but of these, each arrangement is one among $r!\,(n-r)!$ which contain the same elements among the first $r$. So there are $n!/r!(n-r)!$ distinct ways to select a subset of $r$ elements from a set of $n$.
$${^n\mathrm C_r}~=~\dfrac{n!}{r!~(n-r)!}~=~\dfrac{^n\mathrm P_r}{r!}$$
That's all you need to be concerned with: The count of ways to select $r$ items from $n$ is $^n\mathrm C_r$.
So the probability for obtaining $2$ from $3$ red, $1$ from $5$ green, (and $0$ from $2$ blue,) marbles, when selecting $3$ from all $10$ marbles is …$$\dfrac{{^3\mathrm C_2}\,{^5\mathrm C_1}}{^{10}\mathrm C_3}$$
We count the favoured event: ways to select $2$ from a set of $3$, and to select $1$ from a set of $5$.
We count the total space: ways to select $3$ from a set of $10$.
We divide and calculate.