Order of picking items in Combinatorics This may have been answered before, I just can't word my question properly in order to find it...
Given 5 red balls and 8 black balls, we need to pick 9 balls.
Are these two questions equivalent, if so, why?

*

*In how many ways can I firstly pick 3 red balls and then 6
black balls.

*In how many ways can I pick 3 red balls and 6 black balls.

Knowing I can't reuse a ball, and order matters.
My solution to the second question:
Pick 3 places for red balls, $\binom{5}{3} \times3!$
Pick 6 places for black balls $\binom{8}{6} \times 6!$
Thus, $3!\times\binom{5}{3}\times6!\times\binom{8}{6}$.
 A: It all depends on what counts as different.
Perhaps the answer to (1) is $1$ and to (2) is ${3+6 \choose 3}=84$
But I would find that unhelpful, and would prefer the answer to (1) to be
$$5\times 4 \times 3 \times  8 \times 7 \times 6 \times 5 \times 4 \times 3 =\frac{5! \,8!}{(5-3)!(8-6)!} = 1209600$$
which is your answer to the second question, though I think that (2) should  still be ${3+6 \choose 3}$ times this, i.e. $9!{5\choose 3}{8 \choose 6} = 101606400$
This latter approach makes calculating probabilities possible: the total number of ways of choosing $9$ balls this way is $\frac{13!}{(13-9)!}$ making the probability of drawing $3$ red and $6$ black in any order $\dfrac{{5 \choose 3}{8\choose 6}}{13 \choose 9}$, the usual hypergeometric probability.  The first approach would not facilitate this calculation
A: I think you should relate your query to probability rather than number of ways. It is there that I've seen students err.
If the order is specified, P(3 red followed by 6 black) would be
$\frac5{13}\frac4{12}\frac3{11}\frac8{10}\frac79\frac68\frac57\frac46\frac35$
but often, when the draws are to be in an unspecified order, it is mistakenly computed as above, whereas of course it needs to be multiplied by $\frac{9!}{3!6!},$
or better computed as$\dfrac{\binom53\binom86}{\binom{13}9}$
And conversely, when sometimes a specified order is given, the formula
$\dfrac{\binom53\binom86}{\binom{13}9}$ is used whereas, of course, it needs to be divided by $\frac{9!}{3!6!}$
If you ask for the number of ways draws can be made with  specified order and unspecified order, the issue becomes quite murky as you can see from the variety of opinions.

PS:
You write "This may have been answered before,..", I would like to add that I have never seen this asked ever !
