A finishes 1st: 5 options for B x 4 options C x 3 x 2 options for D,E,F = 120
2nd: 4....3...." = 72
3rd: 3.....2..." = 36
4th: 2........." = 12

Total 240 C

Hello GMATinsight and Regor.

I would like to give you a more detailed solution.

We have 6 horses and we can have the following cases:

- Horse A finishes 1st: here, we have to organize 5 horses in 5 places, the number of ways to do that is A*5*4*3*2*1=5!=120.

- Horse A finishes 2nd: First, we have to pick the horse for the first place but we can not select horses B or C. So, we have 3 options for the first place (D, E, F) and the number of ways to select a horse for the first place is 3. Then we have to organize 4 horses in the last 4 places. The total number of ways in this case is 3*A*4*3*2*1=72.

- Horse A finishes 3rd: as in the case before, we have to pick 2 horses for the first 2 places but we can not select horses B or C. So, we have 3 options again and we can organize this 2 horses in 3*2=6 different ways. Now we have to organize the last 3 horses in the last 3 places. The total number of ways in this case is 3*2*A*3*2*1=36.

- Horse A finishes 4th: Now, we have to pick 3 horses for the first 3 places but we can not select horses B or C, the number of ways to organize this 3 horses is 3*2*1=6. Finally, we have to organize 2 horses (B and C) in the last 2 places. The total number of ways in this case is 3*2*1*A*2*1=12.

You should noticed that A can not finish in the 5th place or in the 6th place.

In conclusion, the total number of ways this race can end is 120+72+36+12=240.

So, the correct answer is C.

I hope this help you. I'm available if you'd like any follow up.