Counting Question involving cards I'm preparing for an exam in my stats class and there's a question that I can't seem to get the right answer to. The question says that from a deck of 40 cards (10 suits A-10), 4 are chosen randomly. What is the probability that you get 2 hearts and 2 10s?
My logic is as follows. Since the 10 of hearts satisfies both conditions its a special case. Either its chosen or it isn't. In the case it's chosen, I have to choose it so (1 choose 1)*(3 choose 1) (because I have to pick another 10) and then out of the 9 remaining hearts, I have to chose 1 of them. Then, I'm free to select any of the 37 remaining cards.
If the 10 of hearts isn't chosen, then from 3 other 10s, I have to choose 2 and then from the 9 other heart cards, I choose 2.
In summary, ((1 nCr 1)(3 nCr 1)(9 nCr 1)(37 nCr 1)+(3 nCr 2)(9 nCr 2))/(40 nCr 4). But this doesn't yield the right answer. Where am I going wrong?
 A: I’m assuming that any hand with at least two $10$’s and at least two hearts counts.
In the case in which you choose the $\heartsuit 10$ you’re overcounting the sets of $4$ cards that include more than two hearts. Consider for instance the set $\{\heartsuit 10,\clubsuit 10,\heartsuit 3,\heartsuit 7\}$: you count it once as

pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 3$, pick another card

and once as

pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 7$, pick another card.

You’re also overcounting sets like $\{\heartsuit 10,\clubsuit 10,\heartsuit 3,\spadesuit 10\}$, once as 

pick the $\heartsuit 10$, pick the $\clubsuit 10$, pick the $\heartsuit 3$, pick another card

and once as

pick the $\heartsuit 10$, pick the $\spadesuit 10$, pick the $\heartsuit 3$, pick another card.

There are $\binom{39}3$ sets that contain the $\heartsuit 10$. $\binom{30}3$ of them contain no other heart, and $\binom{37}3$ of them contain no other $10$, so a first approximation to the number of sets that include the $\heartsuit 10$ and at least one other heart and one other $10$ is $\binom{39}3-\binom{30}3-\binom{37}3$. However, this undercounts: each set that contains no other heart and no other $10$ has been subtracted off twice. There are $\binom{27}3$ such sets, so the number of sets that include the $\heartsuit 10$ and at least one other heart and one other $10$ is actually
$$\binom{39}3-\binom{30}3-\binom{37}3+\binom{27}3=234\;.$$
If you’re counting only those hands with exactly two $10$’s and two hearts, your error is in saying that there are $37$ possible choices for the fourth card: there are only $27$, since it can’t be one of the remaining eight hearts or two $10$’s. In this case all you need do to get the right answer is change $37$ to $27$.
In both versions your count of the cases in which the $\heartsuit 10$ is not chosen is fine.
A: The question is imprecisely stated. Does it mean a hand that has exactly two hearts and exactly two $10$'s? Or does it mean at least two hearts and two $10$'s?
$1.$ We first interpret the question as meaning exactly.
Then as you point out there are two cases. If one of the cards is the $10$ of hearts, then the other $10$ can be chosen in $\binom{3}{1}$ ways, and for each such choice the other heart can be chosen in $\binom{9}{1}$ ways. Now the remaining card can be chosen in $\binom{27}{1}$ ways, giving $\binom{3}{1}\binom{9}{1}\binom{27}{1}$ ways.
For the hands that do not have the $10$ of hearts, the count is easier: the $10$'s can be chosen in $\binom{3}{2}$ ways, and and for each choice, the hearts can be chosen in $\binom{9}{2}$ ways, giving $\binom{3}{2}\binom{9}{2}$ ways. Now add. 
$2.$ Now we deal with the interpretation "at least." We can recycle the previous result, by adding to the earlier count the hands that have three $10$'s including the $10$ of hearts, and the hands that have three hearts including the $10$. There are $\binom{3}{2}+\binom{9}{2}$ of these.  
