Remark
There are two possible answers, unfortunately. Perhaps the answer choices will straighten it out.
One
If the first 4 finishers can be in any one of the first 4 places, you have a combination.
That is written as [tex]\dbinom{n}{k}=\dbinom{15}{4} = \dfrac{15!}{11!*4!}[/tex]
which is equal to 1365
Two
But the problem is that the four finishers could be in a specified order. in which case you do not divide by 4! 13760 is your answer.
OK so both answers are there. What do we do. The question does not specify the second possibility as being there. So I would choose 1365 <<< Answer.
But do not be surprised if the answer is 13760. If it is, then respectfully complain politely, but move on.
Note
The first answer comes from 15C4 which means the top 4 can finish in any order. The question is then a combination problem
The Second answer comes from 15P4 which means there is 1 specific order that the top 4 finish in. I do not believe that is what the question means.
