Really simple probability question I'm having trouble calculating some probabilities for a simple card game I'm into 
1) So say we have a deck of 13 cards total. You draw 3 cards so there are 10 cards left in the deck. Say there is 1 special card in the deck that we want to figure out the probability of getting it in the hand. How do we figure it out? Would it be:
$$ \dbinom{1}{1} *  \dbinom{12}{2} /  \dbinom{13}{3} $$
1 choose 1 for the special card. 12 choose 2 for the other 2 cards which we don't care what they are. I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand but that would make the probability greater than 1.
2) Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card? 
3) Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc).
I remember doing math like this in high school and it being really basic but don't quite remember exactly how to solve them. Thanks!
 A: Looks right.
"I was thinking I might have to multiply by 3! because there are 3! ways to order the 3 cards in the hand " --> No need. Order doesn't matter in this case.
"Now let's say in the deck there is that 1 special card, and also 3 copies of a different type of card that we want. How would we calculate the probability of forming a hand that contains the 1 special card AND 1 or more of any of the 3 copies of the 2nd type of card?"
1 special card, 1 of 2nd type
$\binom{1}{1} \binom{3}{1} \binom{9}{1} / same$
1 special card, 2 of 2nd type
$\binom{1}{1} \binom{3}{2} \binom{9}{0} / same$
1 special card, 3 of 2nd type
impossible since we draw 3
"Now let's say we have a deck with 2 copies of card type A, and 2 copies of card type B. How would we calculate the probability of choosing 1 or more from type A AND 1 or more from type B assuming a 13 card deck where we draw 3 cards? (for example: 1 type A + 1 type B + 1 any other card, 1 type A + 2 type B, etc)."
X type A, Y type B
$\binom{2}{X} \binom{2}{Y} \binom{10}{3-X-Y} / same$
where X = 1 or 2, Y = 1 or 2 and X + Y $\leq 3$
A: Hint on 2)
$P(\text{the special AND}\geq1 \text{ copies of...})=P(\text{the special})-P(\text{the special AND 0 copies of...})$
