A practical difficulty that sometimes arises in computing the probability of an event is counting the number of basic outcomes in the sample space and the event of interest.
Suppose that we have some number \(n\) of objects that are to be placed in order. Each object may be used only once. How many different sequences are possible? There are \(n\) ways to choose the first object, there are \(n-1\) ways to choose the second object etc.
So, the total number of possible ways of arranging x objects is
\[n! = n \cdot (n-1) \cdot (n-2) \cdot \cdot \cdot 3 \cdot 2 \cdot 1\] \(n!\) is read as “n factorial” and we define \(0!=1\).
Suppose that we have \(n\) objects. How many ways can we arrange \(x\) of these objects where \(x<n\)? Each object may be used only once. There are \(n\) ways to choose the first object, there are \(n-1\) ways to choose the second object, and so on, until we come to the last object in which there there are \(n-x+1\) ways to choose it which leads us to the following result:
\[P^n_x = n \cdot (n-1) \cdot (n-x+1)=\frac{n!}{(n-x)!}\]
Example 1:
Suppose that two letters are to be selected from A, B, C, D, and E. How many permutations are possible?
Click For AnswerWhat if we are interested in the number of different ways that \(x\) objects can be selected from \(n\) objects where no object may be chosen more than once but are not concerned about the order. Many of the permutations will be rearrangements of the same objects (i.e. AB and BA in the example above). Since \(x\) objects can be arranged in \(x!\) ways, we are only concerned with \(\frac{1}{x!}\) of the permutations.
\[C^n_x = \frac{P^n_x}{x!} = \frac{n!}{(n-x)!x!}\]
Example 2:
A personnel officer has four candidates (Mary, Alice, Joe and Bob) to fill two similar positions. How many ways can the two candidates be chosen?
Click For Answer