Goldenrod and No Hope are in a horse race with 4 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?
Options
A) 720 B) 360 C) 12 D) 24 E) 21
All 4 horses can finish the race in 4! way (assuming no tie).   If no tie is possible between No Hope and Goldenrod, then in half of these cases No Hope will be before Goldenrod and in half of these cases after (not necessarily right before or right after). How else? So, there are 4!/2=12 different arrangements of finishes where No Hope always finishes before Goldenrod.   Answer: 12.
<result>C</result>
Goldenrod and No Hope are in a horse race with 4 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race?