Ways to choose $4$ pieces from $4$ B's and $3$ H's and $2$ S's? 
One plays $4$ pieces out of $9$. $4$ of them are (of type) B, $3$ are H and $2$ are S. How many ways the pieces can be chosen if there must be at least 1 of each type?

I figure out 2 methods.
1 is laying out all possible scenarios:
1 2B, 1H, 1S
2 1B, 2H, 1S
3 1B, 1H, 2S
That gives me $72$.
2nd is choose one each first, then choose $1$ out of $6$ for the remaining i.e.  $C(4,1) · C(3,1) · C(2,1) · C(6,1)$. That gives me $143$.
2nd method yields double the first. I think 2nd method may double counted something but I can't figure out. Can anyone help?
 A: Yes the second is overcounting.
Suppose you chose B1, C1, D1 in the first step then chose B2 in the second step.
In another combination , you chose B2, C1 , D1 in the first step and then later in the second step chose B1.
Though they are counted differently , you ended up with the same set of pieces
Each selection is counted twice, giving a double count
A: You want the coefficient of $x^4$ in
$$(4x+6x^2)(3x+3x^2)(2x+x^2)$$
which is $72$.
Your second method counts each one twice. For example, if the base set was $B_1H_1S_1$ and you added $B_2$, you have the same result as if the base set was $B_2H_1S_1$ and you added $B_1$.
A: First, I assume that the ways pieces can be chosen do not depend of the order of the pieces you take. In that case, the formula of the number of possibilities that 4 pieces can be picked out of 9 is given by "n choose k":$$C_k^n = \frac{n!}{(n-k)!k!}$$
In our case, we have $n=9$ and $k=4$, so $C_4^9 = \frac{9!}{5!4!} = 126$. Of course, we have some restrictions (1 piece of B, H and S), so the answer will be below 126. In this exercice, as you only take 4 pieces and that 3 pieces will belong to different families, it means that you will have one and one only double. Thus, I'll try to calculate the number of combinations you can have with 2 pieces of the same family and the other 2 pieces of different families:

*

*How many 2B, 1H and 1S do I have ? If I pick 2B, I need to have 2 pieces of H and S (so 2 out of 5, given by $C_2^5 = 10$. I have to subtract the number of possibilities that 2S and 2H are picked, so 1 for 2S and 3 for 2H (2+1). I have thus 6 possibilities here, that you need to multiply with the number of different "2B" you can have = 6 (3+2+1), so it's 36.

*How many 2H, 1B and 1S do I have ? $C_2^6 = 15$, 1 possibility for 2S, 6 for 2B (3+2+1). We have 8.3=24 possibilities (2+1).

*How many 2S, 1B, 1H do I have ? $C_2^7$ = 21, 6 ways of having 2B, 3 ways for 2H, so 12 that you multiply by 1 (only 2S available).

That gives you 36+24+12 = 72 possibilities. To conclude, I'd say your second method is problematic since you have more that 126 possibilities, which is not possible.
As I'm not a professional of probabilities, I may have the wrong answer or a too long calculation, and I would be grateful if somebody could confirm what I just did.
