In a horse race with 6 horses, how many possible finishing orders are there (no ties)?
In a horse race with 6 horses, there are 720 possible finishing orders.
The number of possible finishing orders for a set of items can be calculated using the factorial of the number of items. In this case, with 6 horses, the calculation is 6! (6 factorial), which equals 720.
This value represents the factorial of 5 (5!), which corresponds to the number of ways to arrange 5 horses. Since we have 6 horses in this race, the total number of arrangements must include all 6, making 120 an insufficient number of finishing orders.
This choice is equal to 4! multiplied by 6 (4! * 6), which incorrectly suggests that the arrangements of 4 horses and the inclusion of 2 additional positions would yield the total arrangements. However, as we are dealing with 6 horses, this calculation does not reflect the factorial of 6, leading to an incorrect answer.
This number is the result of 6! divided by 2 (6! / 2), which would imply that half of the possible arrangements are being considered. Such a division does not apply to the scenario of counting all finishing orders of 6 horses without ties, making 360 an incorrect representation of the total arrangements.
This is the correct calculation, as it is the result of evaluating 6! (6 factorial), which equals 6 × 5 × 4 × 3 × 2 × 1 = 720. This number accurately represents all the unique ways in which 6 horses can finish a race.
The total number of finishing orders for 6 horses in a race is determined by calculating 6!, which results in 720 distinct arrangements. The other options fail to account for all horses correctly and either underestimate or miscalculate the arrangements, leading to incorrect answers. Thus, understanding the use of factorials is crucial for solving permutation problems like this one.
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