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Unformatted text preview: Chapter 6.4: Advanced Counting Problems Today we will combine all of our counting techniques to tackle more complicated counting problems. We will also learn techniques for arranging objects where some objects must be kept together, and for arranging objects where some objects must be separated. Example: Suppose we wish to arrange some books on a bookshelf. There are four different math books, three different biology books, and six different chemistry books. How many ways are there to arrange the books if there are no restrictions? If there are no restrictions on the ordering of the books, then all we are doing is arranging thirteen distinct objects. This can be done 13! = 6227020800 ways. Example: How many ways are there to arrange the books if we want books of the same subject kept together? To keep the books together, we will imagine that the math books are placed in one box, the biology books in another box, and the chemistry books in a third box: M 1 ,M 2 ,M 3 ,M 4 , B 1 ,B 2 ,B 3 , C 1 ,C 2 ,C 3 ,C 4 ,C 5 ,C 6 To arrange the books so that they are kept together by subject is done in two steps: 1. Arrange the boxes containing the books. 2. Arrange the books inside each box. There are three boxes. They can be arranged in 3! different ways. There are four different books in the box of math books. They can be arranged 4! ways. There are three different books in the box of biology books. They can be arranged 3! ways. There are six different books in the box of chemistry books. They can be arranged 6! ways. 1 We are performing these four tasks (arrange the boxes, arrange the math books, arrange the biology books, arrange the chemistry books) one after another. By the rule of product, there are 3!4!3!6! = 622080 ways to arrange the books so that they are kept together by subject. Example: Suppose that just the math books need to be kept together. How many different arrangements are there? Since the math books are the only ones that need to be kept together, we will imagine that they are placed inside a box, but that the biology and chemistry books are not in a box: M 1 ,M 2 ,M 3 ,M 4 ,B 1 ,B 2 ,B 3 ,C 1 ,C 2 ,C 3 ,C 4 ,C 5 ,C 6 We have a total of ten objects (9 individual books and 1 box of books). These 10 objects can be arranged in 10! ways. At this point, we have only decided on the order of the individual books and the box of books. We have not yet decided on an ordering for the math books inside the box. These four books can be arranged in 4! different ways inside the box. Since the two tasks (arrange the individual books and box of books, arrange the books inside the box) are done one after the other, then we use the rule of product. There are 10!4! = 87091200 ways to arrange the books so that the math books are kept together....
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 Spring '11
 stephenlang
 Calculus, Counting

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