elemprob-fall2010-page11

elemprob-fall2010-page11 - / 2! = 12. If there were 3 a s,...

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An example. What is the number of possible batting orders with 9 players? Answer: 9! An example. How many ways can one arrange 4 math books, 3 chemistry books, 2 physics books, and 1 biology book on a bookshelf so that all the math books are together, all the chemistry books are together, and all the physics books are together. Answer: 4!(4!3!2!1!). We can arrange the math books in 4! ways, the chemistry books in 3! ways, the physics books in 2! ways, and the biology book in 1! = 1 way. But we also have to decide which set of books go on the left, which next, and so on. That is the same as the number of ways of arranging the letters M,C,P,B , and there are 4! ways of doing that. How many ways can one arrange the letters a,a,b,c ? Let us label them A,a,b,c . There are 4!, or 24, ways to arrange these letters. But we have repeats: we could have Aa or aA . So we have a repeat for each possibility, and so the answer should be 4!
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Unformatted text preview: / 2! = 12. If there were 3 a s, 4 b s, and 2 c s, we would have 9! 3!4!2! . An example. Suppose there are 4 Czech tennis players, 4 U.S. players, and 3 Russian players, in how many ways could they be arranged? Answer: 11! / (4!4!3!). What we just did was called the number of permutations. Now let us look at what are known as combinations. How many ways can we choose 3 letters out of 5? If the letters are a,b,c,d,e and order matters, then there would be 5 for the rst position, 4 for the second, and 3 for the third, for a total of 5 4 3. But suppose the letters selected were a,b,c . If order doesnt matter, we will have the letters a,b,c 6 times, because there are 3! ways of arranging 3 letters. The same is true for any choice of three letters. So we should have 5 4 3 / 3!. We can rewrite this as 5 4 3 3! = 5! 3!2! 11...
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