With a short time remaining in the day, a fedEx driver has time to make deliveries at three locations among the eight locations remaining. How many different routes are possible? did i do it right 8/3= 2.67
To determine the number of different routes possible for the FedEx driver, we need to use the concept of combinations. In this case, we want to find out how many ways the driver can select three locations out of the remaining eight locations.
The formula for calculating combinations is nCr, where n is the total number of items and r is the number of items being selected at a time. To calculate nCr, we need to use the formula:
nCr = n! / (r!(n-r)!)
In our case, n = 8 (total number of remaining locations) and r = 3 (number of locations the driver can make deliveries to). Plugging these values into the formula, we have:
8C3 = 8! / (3!(8-3)!)
Simplifying further:
8C3 = 8! / (3! * 5!)
Next, we need to calculate the factorials. A factorial of a number is the product of all positive integers less than or equal to that number.
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
3! = 3 * 2 * 1
5! = 5 * 4 * 3 * 2 * 1
Now, substitute the factorials back into the formula:
8C3 = (8 * 7 * 6 * 5 * 4 * 3 * 2 * 1) / [(3 * 2 * 1) * (5 * 4 * 3 * 2 * 1)]
Simplifying further:
8C3 = (8 * 7 * 6) / (3 * 2 * 1)
8C3 = 336 / 6
8C3 = 56
Therefore, there are 56 different routes possible for the FedEx driver to make deliveries at three locations among the remaining eight locations.
So, to answer your question, no, 8/3 = 2.67 is not the correct calculation. The correct calculation is 8C3 = 56, as explained above.
To determine the number of different routes possible, you need to calculate the combinations of selecting 3 locations out of the remaining 8.
The formula to calculate combinations is:
C(n, r) = n! / (r! * (n - r)!)
Where:
n is the total number of items (locations in this case),
r is the number of items to be selected (number of locations where deliveries can be made),
Using the formula, we can calculate the number of different routes:
C(8, 3) = 8! / (3! * (8 - 3)!)
= 8! / (3! * 5!)
= (8 * 7 * 6 * 5!) / (3 * 2 * 1 * 5!)
= (8 * 7 * 6) / (3 * 2 * 1)
= 56
Therefore, there are 56 different routes possible for the FedEx driver to make deliveries at three locations among the remaining eight.
routes depend on connecting roads, but there are
8C3 = 56 ways to pick 3 from 8 items.
There are, however, 8P3 = 336 different ways those 3 items can be arranged.
The idea of 2.67 different routes is meaningless. What is a fraction of a route?