Ranking athletes by nationality In an international track competition, there are five United States athletes, four Russian athletes, three French athletes, and one German athlete. How many rankings of the 13 athletes are there when
b) Only nationality is counted and all the U.S athletes finish ahead of all the Russian athletes?
Would the answer by C(13, 9) * P(8; 4, 3, 1)?
 A: HINT: Rankings for this competition are simply arrangements of the 13 athletes. Since only nationality is counted, there are $\frac{13!}{5!4!3!}$ possible rankings (and 13 possible ranks). There are $5!$ possible rankings among the Americans, and $4!$ among the Russians. How can you find the number of possible rankings if all the U.S. athletes beat all the Russian athletes?       My advice is to simplify the problem and then generalize the mechanism (this helps with a lot of combinatorics problems). In other words, change the problem so that you have, say, 7 possible ranks, and fewer athletes. And think of the ranks as boxes into which you can put the athletes, subject to the constraint that some of them (the American ones) are always ahead of others (the Russian ones). After you do this, you should discover how it works, and you can then apply it to your original problem.
A: Line up the 5 Americans and the 4 Russians.  We can think of these 9 people as dividers, and then there are $\binom {12}{3}$ ways to distribute the 3 French athletes between these dividers.  This leaves 13 gaps into which the German can be placed, so the answer should be $13\binom{12}{3}$.
Here's another approach:  
Color 3 balls red, 1 ball black, and 9 balls green;
these can be arranged in $\displaystyle\frac{13!}{9!3!}$ ways. (The first 5 green balls correspond the Americans and the remaining 4 green balls correspond to the Russians.)
A: Think of it as the number of ways of arranging 13 balls, 3 of which are green, 5 are red, 4 are blue and 1 is white, such that none of the blue balls come before any red ball.
this is 4[C(8,4) + 5.C(7,4) + C(6,4).C(6,4) + C(7,3).C(5,4) + C(8,4)] .
