Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP. Bhisham C. Gupta

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 and are B's. The number of distinguishable arrangements of these n letters (x A's and B'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 B's in n places.

      The number is usually called binomial coefficient, since it appears in the binomial expansion (for integer )

      (3.4.5)

      Example 3.4.5 The coefficient of in the expansion of is , since we can write as

      (3.4.6)

      The coefficient of is the number of ways to pick x of these factors and then choose a from each factor, while taking b from the remaining factors.

      3.4.4 Arrangements of n Objects Involving Several Kinds of Objects

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 Скачать книгу