Number of ways to rank players in a tournament Six players play chess, and there's one two-way tie. How many distinct final arrangements are possible?

I think there are ${6\choose 2}$ ways to choose the two players who will tie. Then, there are $5!$ ways to rank the five players (we can think of the two people who tied as one person). So is the answer just ${6\choose 2} \cdot 5!$?

Yes, that's correct. 
Another way to get the same result is to start with the $6!$ rankings of all 6 players, then declare which 2 players with neighbouring ranking are actually in the tie. That's 5 possibilities (1 and 2, 2 and 3, ... 5 and 6). Then we need to take into acccount that each tie is reached exactly 2 times (order "1,2,3,4,5 ties with 6" is reached from initial orders "1,2,3,4,5,6" and "1,2,3,4,6,5", for example).
This yields the same result
$$6!\times5\times\frac12 = {6 \choose 2} 5!$$
