# Counting ways when first position is covered with a tie

Six horses are entered in a race. If two horses are tied for first place and there are no ties among the other four horses, in how many ways can the six horses cross the finish line?

My approach:
We choose $$2$$ out of $$6$$ for the first place and $$4$$ out of the remaining $$4$$:
$$_6P_2\cdot _4P_4 = \frac{6!}{(6 -2)!}\cdot\frac{4!}{(4 - 4)!}=720$$

But the solution states $$360$$.

What am I doing wrong here?

• Also, $\binom 44= \frac{4!}{0!4!}$. But you don't want $\binom 44$ because you care about the order of the remaining $4$ horses. You want the number of ways to permute those $4$ horses, which is simply $4!$. Feb 26 at 23:31
• @RobertShore: Yes the order matters that is why I used $\binom{n}{k}=\frac{n!}{(n-k)!}$ which is $n\cdot(n-1)\cdots\cdot(n-k+1)$. Also it is still $720$
– Jim
Feb 26 at 23:35
• Your values for the combinations are wrong. ${6 \choose 2}=\frac {6!}{4!2!}=15$. You don't want $4 \choose 4$ because the order of the last four horses matters. That gives $15 \cdot 4!=360$ Feb 26 at 23:40
• When you write $6 \choose 2$ you are saying that order does not matter. I don't know a simple one for when order does matter. Some use $_6P_2$ for that. Feb 27 at 0:09
• There are $\binom{6}{2}$ to select the subset of two horses which finish in a tie and $4!$ ways for the remaining four horses to finish in the remaining four positions. Notice that the order of selection for the two horses which finish in a tie does not matter, which is why your answer $P(6, 2)P(4, 4)$ is twice the correct answer. Feb 27 at 23:28

That leaves 4 choices for the horse to come in third, three choices for the horse to come in fourth, two choice for the horse to come in fifth, and one for the horse to come in sixth. The number of ways this can happen is $$15(4)(3)(2)(1)= \frac{6!}{2}= 720/2= 360$$,