How many combinations of winners will there be? 
In a weight-lifting competition, there are 3 Americans, 4 Russians, 2 Chinese, and 1 Canadian. How many outcomes will there be such that 1 American is among the top three and 2 Americans will be among the bottom three?

What I said was that, because there is only one American in the first three, then I must choose two other competitors from seven in order to fill out the other two slots $\binom{7}{2}$. I must do the same thing for the bottom slot not filled out by the Americans yet, therefore I get $\binom{5}{1}$. Multiplying both of them by three (for the different permutations of the places among themselves) I get $$9\binom{7}{2} \binom{5}{1} = 945$$ Which according to the book is the correct answer, but what happened to the other four slots? Should I not multiply that final answer by $4!$ to get the actual outcome? How else might I do this problem? 

Apparently, only  nationalities are to be considered, not individuals (strange though it is !)
[May be, just like medal counts in the Olympics, where only countries figure in the tallies ! ]
The $3$ Americans can be positioned in $\binom31\binom32$ = 9 ways,
and the rest positioned in $\dfrac{7!}{4!2!1!}$ ways.
Multiply the two to get the ans of $\;945$
