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 $3!4!$?

  • $\begingroup$ Are the dogs distinguishable? $\endgroup$
    – Bram28
    Dec 13 '17 at 19:47
  • $\begingroup$ I believe that was the author's intention, yes. $\endgroup$
    – Peetrius
    Dec 13 '17 at 20:22

Once we have ordered the first three dogs, it doesn't really matter to us whether the remaining animals are dogs or cats.

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.


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