Trying to understand how to count with more than 1 group of balls Let us assume that we have 5 balls(blacks have letters: a, b, c, and whites are d and e).
How many possibilities do we have for taking 1 black and 1 white(no replacement, order is important)? I would say $3\times 2\times 2!$, where the 2! is there since we could take (3b,2w) or (2w, 3b).
How many possibilities do we have for taking 1 black and 2 white(no replacement, order is important)? I would say $3\times 2\times 1 \times \frac{3!}{2!}$, since we have (3b,2w,1w) or (3b,1w,2w) as the same, but different from (2w,3b,1w) for example, and so we divide by $2!$.
How many possibilities do we have for taking 1 black and 2 white(no replacement, order is not important)? I think the answer is supposed to be $\frac{3}{1!} \times \frac{2\times 1}{2!}=C^3_1C^2_2$. The $3!$ disappeared since now we have (3b,2w,1w), (1w,3b,2w), (2w,1w,3b) as the same, for example, and so had to divide by $3!$ .
Edit: Now we have 30 balls:  7 blacks, 3 whites, 15 yellows, 5 greens.
Number of possibilities, in 5 draws, of 2b, 3y (no replacement, order important)? $\frac{7\times 6}{2!}\times \frac{15 \times 14 \times 13}{3!} \times 5! $
Number of possibilities, in 5 draws, of 2b, 3y (no replacement, order isn't important)? $\frac{7\times 6}{2!}\times \frac{15 \times 14 \times 13}{3!}=C^7_1 C^{15}_3 $
Number of possibilities, in 6 draws, of 2b, 3y,1g (no replacement, order is important)? $\frac{7\times 6}{2!}\times \frac{15 \times 14\times 13}{3!} \times\frac{5}{1!}\times 6! $
Are my calculations correct?
 A: You clarified in comments that when you said,

...we have (3b,2w,1w) or (3b,1w,2w) as the same, but different from (2w,3b,1w) for example, and so we divide by $2!$,

you meant that the color labels, BWW, are the same for the first two, but different for the last one, WBW. I agree with all your answers, and I agree, if I understand you correctly, that you divide by $2!$ because W appears twice as a color label.
One observation: when you write

number of possibilities, in 5 draws, of 2b, 3y (no replacement, order important)? $\frac{7×6}{2!}×\frac{15×14×13}{3!}×5!$,

we can interpret this as $\binom{7}{2}\binom{15}{3}\cdot5!$, or as the number of ways of carrying out the process

*

*choose $2$ black balls;

*choose $3$ yellow balls;

*arrange the $5$ selected balls in some order.

An alternative expression is
$$
\frac{5!}{2!\,3!}\times(7\cdot2)\times(15\cdot14\cdot13),
$$
which can be interpreted as $\binom{5}{2,3}P(7,2)P(15,3)$, or the number of ways of carrying out the process

*

*label $5$ slots with color labels B, B, Y, Y, Y;

*select and arrange balls in the B slots;

*select and arrange balls in the Y slots.

Interestingly, the second method can be applied to the situation where the balls are replaced. In this case you get
$$
\frac{5!}{2!\,3!}\times7^2\times15^3.
$$
This is similar to a problem that was in an earlier version of your post.
Added in response to comment: I regard it as a mistake to say that you divide by $2!$ to consider $(a,d,e)$ and $(a,e,d)$ the same. There are two ways to think about what the $2!$ is really for, depending on which of the two process above you are using.
If you are using the first process, there are $\frac{3}{1}$ sets containing one black ball, $\frac{2\cdot1}{2!}$ sets containing two white balls (there's your division by $2!$), and $3!$ ways of arranging the elements of a three-element set formed by taking the union of a set of one black ball and a set of two white balls. When you divide by $3!$ to go to the unordered version of the problem, you are simply undoing the final arrangement step. You could say that you divided by $2!$ because the sets $\{d,e\}$ and $\{e,d\}$ are the same set, and not because we want the ordered triples $(a,d,e)$ and $(a,e,d)$ to be considered the same. (Dividing by $3!$ takes care of the latter.)
If you are using the second process, there are $\frac{3!}{2!}$ ways to put the color labels B, W, W on the slots (there's your division by $2!$), $3$ ways to put a black ball in the B slot, and $2\cdot1$ ways to put white balls in the two W slots. To do the unordered version of the problem, you divide by $3!$ to turn the number of ordered triples into the number of sets. You could say that you divided by $2!$ because the color labelings BWW and BWW are the same labeling. (Normally there are $3!$ permutations of three labels, but when two of the labels are identical, $3!$ overcounts the true number of permutations by a factor of $2!$.)
