# Permutations of finishing a race

Four cats and six dogs are in a race. How many ways exist if a dog must come first, second, and third.

If a dog must come in the first three positions of a ten position race then the number of ways the first three positions can be ordered is $\frac{6!}{(6-3)!}$ or 120 ways. My question is are the next seven positions expressed as $7!$ or $3!4!$?

• Are the dogs distinguishable? Dec 13 '17 at 19:47
• I believe that was the author's intention, yes. Dec 13 '17 at 20:22

So once we've ordered the leaders in $_6P_3 = 120$ ways, we can order the rest in $7!$ ways, for a total of $_6P_3 \cdot 7! = 604800$ ways.