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 ﬁrst 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 doesn’t 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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This note was uploaded on 12/29/2011 for the course MATH 316 taught by Professor Ansan during the Spring '10 term at SUNY Stony Brook.
 Spring '10
 ansan
 Math

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