On each business trip, a salesman visits three of the eight major cities in his territory. In how many different ways can he schedule his route (that is, the cities and their order) for each trip?
First you need to determine if it's permutation or combination. If it were a permutation, the salesman would permute (def permute: to change the order or arrangement of; especially to arrange in all possible ways) his 4 visits. That means he would be bouncing back and forth between the cities, which I'm sure is not the case. So it's a combination. He only visits a city once per trip.
Now for the actual number. Since it's a combination, you can use the nCr function.
12 nCr 4 = 12!/(4!*8!) = 495
The function is literally pronounced "12 choose 4"
I read it differently.
First we have to choose (combination) 3 of the 8 cities, which is C(8,3) or 8C3
= 56
(where did you get 12 cities ?)
now for each of those choices, he can schedule the order of visit (permutation).
Each choice can be arranged in 3! or 6 ways.
number of ways = 56x6 = 336
To solve this problem, we can use the concept of combinations. In this case, we need to find the number of ways the salesman can choose 3 cities out of 8 major cities to visit.
The formula to calculate combinations is given by:
C(n, r) = n! / (r! * (n - r)!)
Where n represents the total number of items, and r represents the number of items to be selected.
In this case, n = 8 (the total number of cities) and r = 3 (the number of cities to be selected).
So, we can substitute the values into the formula:
C(8, 3) = 8! / (3! * (8 - 3)!)
Calculating the factorial:
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 40320
3! = 3 * 2 * 1 = 6
5! = 5 * 4 * 3 * 2 * 1 = 120
Substituting the values:
C(8, 3) = 40320 / (6 * 120)
Cancelling out the common factors:
C(8, 3) = 3360
Therefore, there are 3360 different ways the salesman can schedule his route for each trip.