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Suppose that a collection of n objects are such that there are 's, 's, …, 's, where . Then total number of distinguishable arrangements of these several kinds of A's denoted by is

(3.4.7)

For if we think of each of the n places being originally filled with objects of type A, there are ways of choosing 's to be replaced by 's. In each of these ways, there are ways of choosing 's to be replaced by 's. Hence, the number of ways of choosing 's and replacing them with 's and choosing from the remaining A's and replacing them with 's is . Continuing this argument and using equation (3.4.4) shows that the number of ways of choosing 's and replacing them with 's, A's and replacing them with 's, and so on until the last A's replaced with 's, is

To illustrate the application of combinations to probability problems involving finite sample spaces, we consider the following example.

Example 3.4.6 (Combinations and probability) If 13 cards are dealt from a thoroughly shuffled deck of 52 ordinary playing cards, the probability of getting five spades is

Solution: This result holds because the number of ways of getting five spades from the 13 spades in the deck is , and the number of ways of getting 8 nonspades from the 39 nonspades in the deck is , and hence, the number of ways five spades and eight nonspades occurs in a hand of 13 cards is the product . This is the number of elements in the sample space constituting the event of “getting five spades in dealing 13 cards from a shuffled deck.” Since the sample space consists of equally likely sample points, each sample point is assigned the same probability . Hence, the probability of getting five spades in dealing 13 cards is