Let’s refresh the definitions.
“A permutation of a set of distinct objects is an ordered arrangement of these objects. We also are interested in ordered arrangements of some of the elements of a set. An ordered arrangement of r elements of a set is called an r-permutation. ”
“An r-combination of elements of a set is an unordered selection of r elements from the set. Thus, an r-combination is simply a subset of the set with r elements.”
Both definitions are from the textbook Discrete Mathematics and its Applications by Rosen.
Notice, the exercise state that there is an order in how we must choose the name of the candidates. Therefore, it should be obvious that we need to use the concept of permutations and not that of combinations.
This distinction is the key to answering this (and other) type of exercise.
P(6,6)=6!/(6-6)!
=6!/0!
=6!
=720
Another approach for solving this exercise is as follows.
We can choose the first name in the ballot in 6 different ways (as there are 6 different names). For each of the 6 ways we choose the first name, we can choose the second name in 5 ways, and so on.
It follows that can apply the multiplication rule.
6x5x4x3x2x1 = 720.
Related exercises:
- A group contains n men and n women. How many ways are there to arrange these people in a row if the men and women alternate?
- In how many different orders can five runners finish a race if no ties are allowed?
- Find the value of each of these quantities a) C(5,1) b) C(5,3) c) C(8,4) d) C(8,8) e) C(8,0) f) C(12,6)
- Find the number of 5-permutations of a set with nine elements
- Find the value of each of these quantities
- How many permutations of {a, b, c, d, e, f, g} end with a?
