Combinatorics Example Problem Ten weight lifters are competing in a team weightlifting contest. Of the lifters, 3 are from the United States, 4 are from Russia, 2 are from China, and 1 is from Canada. 
Part 1 
If the scoring takes account of the countries that the lifters represent, but not their individual identities, how many different outcomes are possible from the point of view of scores? 
This part I understand:  10! / [ 3! 4! 2! ]  = 12,600
Part 2 (don't understand this part) 
How many different outcomes correspond to results in which the United States has 1 competitor in the top three and 2 in the bottom three?
This part I'm confused.  Here you have 10 slots. 
The first three slots must be of some order:  US, US, or [Russia, China, Canada]. 
The last three slots must be of some order US, [Russia, China, Canada], [Russia, China, Canada].
I thought the answer would be this:
$\binom{3}{2} \binom{1}{1} * \frac{7!}{4!\ 3!\ 2!} $
My reasoning:  In the first 3 slots, you have to pick 2/3 US people.  Then you only have one remaining.  You have 7! ways to organize the rest but have to take out repeats so you divide by factorials of repeats which is 4,3, and 2.  But my middle term is wrong....
My research shows me answers of 2 forms but I can't understand it: 
Method 1:  $\binom{3}{1} \binom{3}{2} * \frac{7!}{4!\ 3!\ 2!}$
This case, I don't understand why there's a 3 in the second binomial, $\binom{3}{2}$.  We already selected ONE US person so shouldn't it be $\binom{2}{2}$?  
Method 2: $ \binom{7}{2} \binom{3}{1} \binom{5}{4} \binom{3}{2} \binom{1}{1} $
?
Sorry for the long post.  Thanks again.

For the second, you pick the slots (not the people-we said all the people from one country were interchangeable) for the two US people in the bottom ${3 \choose 2}$ ways, but then have to pick which slot the US person in the top is in, which adds  a factor ${3 \choose 1}$.  Then of the seven left, you just have $\frac {7!}{4!2!}$ as there are no Americans among them.  So Method 1 is off by the $3!$ in the denominator.  I don't understand the terms in Method 2.
Added:  One way to look at the first part, which I think is the way you did, is to say there are $10!$ orders of the people, but if the Americans are interchangeable you divide by $3!$, for the Russians you divide by $4!$ and for the Chinese you divide by $2!$.  Another way would be to say we pick three slots for Americans in ${10 \choose 3}$ ways, then pick slots for the Russians in ${7 \choose 4}$ ways, then pick slots for the Chinese in ${3 \choose 2}$ ways, giving ${10 \choose 3}{7 \choose 4}{3 \choose 2}=120\cdot 35 \cdot 3=12600$.  Of course the answer is the same, but the approach is better suited to the second part.  Now if you think of choosing slots for nationalities you get what I did before.
