In a horse race with 6 distinct horses, two horses are named Goldenrod and No Hope. Assuming all 6 horses finish the race with no ties, in how many different possible finishing orders will No Hope finish somewhere ahead of Goldenrod?

Introduction / Context:
This is a classic relative order problem involving permutations with a condition. We have 6 horses and are interested in the finishing order, but with the added restriction that one specific horse, No Hope, must finish before another specific horse, Goldenrod. These types of questions help build intuition for symmetry in permutations and relative ordering constraints.

Given Data / Assumptions:

• Total horses: 6, all distinct.
• Two special horses: No Hope (N) and Goldenrod (G).
• All horses finish the race with no ties.
• We consider all possible full finishing orders of the 6 horses.
• We only count those orders where N finishes somewhere before G.

Concept / Approach:
Without any restriction, all 6 horses can finish in any order, giving 6! possible permutations. Now, observe symmetry: in the set of all possible permutations, the relative positions of N and G are equally likely to be N before G or G before N. Therefore, exactly half of all permutations will have N before G, and the other half will have G before N. This symmetry argument leads to a very quick solution.

Step-by-Step Solution:
Step 1: Compute the total number of unrestricted finishing orders: 6!. Step 2: 6! = 720. Step 3: In half of these permutations, No Hope finishes before Goldenrod. Step 4: In the other half, Goldenrod finishes before No Hope. Step 5: Therefore, number of permutations with No Hope ahead of Goldenrod = 720 / 2. Step 6: Compute 720 / 2 = 360. Step 7: Hence, 360 finishing orders satisfy the condition.

Verification / Alternative check:
As an alternative approach, you could treat N and G as a pair in each permutation and note that for any arrangement of the remaining 4 horses, there are exactly two ways to place N and G in the chosen positions: one where N is earlier and one where G is earlier. Thus, for every arrangement of the 4 other horses and fixed pair positions, half the arrangements have N ahead of G. This again shows that exactly half of the 720 total permutations satisfy the condition, confirming the answer 360.

Why Other Options Are Wrong:
720: This is the total number of orders without any condition, counting both N before G and G before N. 120: This is 5!, not connected to the current situation where 6 horses participate. 640: A random number that does not emerge from any symmetric or factorial based reasoning for 6 horses. Only 360 correctly represents half of the total possible orders.

Common Pitfalls:
A common mistake is to attempt to count all permutations where N is before G directly without using symmetry, which can be complex and error prone. Another pitfall is to divide by an incorrect factor like 6 or 3 instead of 2, misinterpreting the relative positions. Remember that for two distinct labelled items in random permutations, each has an equal chance of being ahead of the other, giving a clean 1/2 probability and simplifying the count.

Final Answer:
The number of finishing orders in which No Hope finishes ahead of Goldenrod is 360.