With no ties we have 3,628,800 permutations (10!).

With one tie (between two racers only) we have 16,329,600 positions. That is (ways of choosing a pair) * (ways of arranging the remaining racers) * (number of positions the tie can be in). That is 9*45*40320.

With one three-way tie we have 8*120*5040 that is 4,838,400.

With one four-way tie we have 7*210*720 that is 1,058,400.

And so it continues.

With two two person ties we have (ways of choosing a pair) * (ways of choosing a pair from the remaining racers) * (number of positions the first pair can be in) * (number of positions the second pair can be in) *(ways of arranging the remaining racers). That is 45*28*8*7*720. That's 50,803,200.

I could continue this but it's obvious that this isn't the method that the problem is looking for (far too much legwork).