College Algebra Exam Review 49

College Algebra Exam Review 49 - 1.9 COUNTING 59 ÂÃ nŠ n...

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Unformatted text preview: 1.9. COUNTING 59 Âà nŠ n The formula D implies (b) when 0 Ä k Ä n. But k kŠ.n k /Š when k is not in this range, neither is n k , so both sides of (b) are zero. When k is 0, both sides of (c) are equal to 1. When k is negative or greater than n, both sides of (c) are zero. For 0 < k Ä n, (c) follows from a combinatorial argument: To choose k elements out of f1; 2; : : : ; ng, we can either choose n, together with k 1 elements out of f1; 2; : : : ; n 1g,  à n1 which can be done in ways, or we can choose k elements out of k1  à n1 f1; 2; : : : ; n 1g, which can be done in ways. I k Âà n Example 1.9.4. The coefficients have an interpretation in terms of k paths. Consider paths in the .x; y/–plane from .0; 0/ to a point .a; b/ with nonnegative integer coordinates. We admit only paths of a C b “steps,” in which each step goes one unit to the right or one unit up; that is, each step is either a horizontal segment from an integer point .x; y/ to .x C 1; y/, or a vertical segment from .x; y/ to .x; y C 1/. How many such paths are there? Each path has exactly a steps to the right and b steps up, so a path can be specified by a sequence with a R’s and b U’s. Such a sequence is determined by choosing the positions of the a R’s, so the number of  Ã à aCb aCb sequences (and the number of paths) is D . a b Âà n The numbers are called binomial coefficients, because of the folk lowing proposition: Proposition 1.9.5. (Binomial theorem). Let x and y be numbers (or variables). For n 0 we have n X Ânà n .x C y/ D xk yn k : k k D0 Proof. .x C y/n is a sum of 2n monomials, each obtained by choosing x from some of the n factors, and choosing y from the remaining factors. For fixed k , the number of monomials x k y n k  à sum is the number of in the n ways of choosing k objects out of n, which is . Hence the coefficient k Âà n of x k y n k in the product is . I k ...
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