Answer
In $90$ ways.
Work Step by Step
The order in which the horses finish the race matters and one horse can not finish in two different positions.
Permutation of 10 distinct horses taken 2 at a time:
$_2P_{10}=\frac{10!}{(10-2)!}=\frac{10!}{8!}$
But, $10!=10\times9\times(8\times7\times6\times5\times4\times3\times2\times1)=10\times9\times8!$
$_2P_{10}=\frac{10\times9\times8!}{8!}=10\times9=90$