Download Algebraic Geometry Santa Cruz 1995, Part 2 by Kollar J., Lazarsfeld R., Morrison D. (eds.) PDF

By Kollar J., Lazarsfeld R., Morrison D. (eds.)

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Extra resources for Algebraic Geometry Santa Cruz 1995, Part 2

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All kelement subsets of A are divided into two disjoint groups according to whether or not these contain x. There are (n–1 k–1) subsets containing x because besides x, we are free to pick (k – 1) elements from the (n – 1)-element set A – {x}. Similarly there are (n–1 k ) subsets not containing x; hence: Using Pascal’s formula and the symmetry of the binomial coefficients, we can form a larger Pascal’s triangle (Fig. 3): NOTES: Let us mention a few facts from the history of Pascal’s triangle and the binomial theorem.

1 + bx + b2x 2 + . )( 1 + c x + c2 x2 +. . ) To see their numbers, set a = b = c = 1 and observe the corresponding coefficients. In general the generating function for the number of k-combinations with repetition of an n-element set is – C ( x ) = ( l + x + x 2 + . . )n = fl 1 = (1- x )– n (1 –x) n As we mentioned earlier, according to Maclaurin’s formula, we have That is, 48 Chapter2 We could have expected this result from our earlier derivations. 38. Let us try a small experiment again. What kinds of problems can be solved by the following generating function: (1 +x +x 2 +x 3 +x 4 +x 2 +x 6 + .

16. How many ordered k-tuples can be formed from the elements of the set A = {a1, a2 , .. , an} if repetition is allowed? 17. How many k-element subsets does the n-element set A have ? SOLUTION: Assume without any loss of generality that the set A is A = { 1,2, . . ,n}. 9, there are as many k-element subsets of A as there are increasing sequences made from its elements. Let the final answer be M. From each of M increasing length-k sequences, we can form k(k – 1). . 1 = k! ordered k-tuples. Hence the number of all ordered k-tuples whose components are different elements of A equals k!

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