The number of sample points in a sample space is equal to the number of branches corresponding to the last trial. For instance, in the present example, the number of sample points in the sample space is equal to the number of branches corresponding to the third trial, which is 24 (). To list all the sample points, start counting from o along the paths of all possible connecting branches until the end of the final set of branches, listing the sample points in the same order as the various branches are covered. The sample space S in this example is
The tree diagram technique for describing the number of sample points is extendable to an experiment with a large number of trials, where each trial has several possible outcomes. For example, if an experiment has n trials and the ith trial has possible outcomes (), then there will be branches at the starting point o, branches at the end of each of the branches, branches at the end of the each of branches, and so on. The total number of branches at the end would be , which represents all the sample points in the sample space S of the experiment. This rule of describing the total number of sample points is known as the Multiplication Rule.
3.4.2 Permutations
Suppose that we have n distinct objects . We can determine how many different sequences of x objects can be formed by choosing x objects in succession from the n objects where . For convenience, we may think of a sequence of x places that are to be filled with x objects. We have n choices of objects to fill the first place. After the first place is filled, then with objects left, we have choices to fill the second place. Each of the n choices for filling the first place can be combined with each of the choices for filling the second place, thus yielding ways of filling the first two places. By continuing this argument, we will see that there are ways of filling the x places by choosing x objects from the set of n objects. Each of these sequences or arrangements of x objects is called a permutation of x objects from n. The total number of permutations of x objects from n, denoted by , is given by
(3.4.1)
Note that the number of ways of permuting all the objects is given by
(3.4.2)
where is read as n factorial.
Expressed in terms of factorials, we easily find that
(3.4.3)
3.4.3 Combinations
It is easy to see that if we select any set of x objects from n, there are ways this particular set of x objects can be permuted. In other words, there are permutations that contain any set of x objects taken from the n objects. Any set of x objects from n distinct objects is called a combination of x objects from n objects. The number of such combinations is usually denoted by . As each combination of each x objects can be permuted in ways, these combinations give rise to permutations. But this is the total number of permutations when using x objects from the n objects. Hence, , so that
(3.4.4)
Example 3.4.2 (Applying concept of combinations) The number of different possible hands of 13 cards in a pack of 52 ordinary playing cards is the number of combinations of 13 cards from 52 cards, and from (3.4.4) is
Example 3.4.3 (Applying concept of combinations) The number of samples of 10 objects that can be selected from a lot of 100 objects is
Example 3.4.4 (Determining number of combinations) Suppose that we have a collection of n letters in which x are A's andare B's. The number of distinguishable arrangements of these n letters (x A's andB's) written in n places is.
We can think of all n places filled with B's, and then select x of these places and replace the B's in them by A's. The number of such selections is . This is equivalent to the number of ways we can arrange x A's and
3.4.1.
