At a certain event, 30 people attend, and 5 will be chosen at random to receive door prizes. The prizes are all same, so the order in which the people are chosen does not matter. How many different groups of five people can be chosen?
Solution
Since the order of the chosen people does not matter, we need to compute the number of combinations of 5 chosen from 30. This is
(30¦5) = 30!/5!25!
= (30)(29)(28)(27)(26)/(5)(4)(3)(2)(1)
= 142,506
Choosing a combination of k objects from a set of n divides the n object into two subsets: the k that were chosen and the n-k that were not chosen. Sometimes a set is to be divided up into more than two subsets. For example, assume that in class of 12 students, a project is assigned in which the students will in groups. Three groups are ways in which the groups can be formed as follows. We consider the process of dividing the class into three groups as a sequence of two operations. The first operation is to select a combination of 5 students to comprise the group of 5. The second operation is to select a combination of 4 students from the remaining 7, to comprise the group of 4. The group of 3 will then automatically consist of the 3 students who are left.
The number of ways to perform the first operation is
(12¦5) = 12!/5!7!
After the first operation has been performed, the number of ways to perform the second operation is
(7¦4) = 7!/4!3!
The total number of ways to perform the sequence of two operations is therefore
12!/5!7! 7!/4!3! = 12!/5!4!3!
= 27,720