exam-1-091001-solutions

exam-1-091001-solutions - MATH 681 Exam #1 1. (8 of 12...

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MATH 681 Exam #1 1. (8 of 12 students attempted this) Prove the combinatorial identity n k =1 k 2 ( n k ) = n 2 n - 1 + n ( n - 1)2 n - 2 . You may use any method you like. A simple approach is purely combinatorial: the left side of the equation represents the number of ways to choose a subset of size k from an n -element set (which can be done in ( n k ) ways)and then select, with order, an element and then another (possibly identical) element of the subset (which can be done in k 2 ways). These numbers are added up over all values of k , so that the left side represents the number of ways to choose a subset of any size, and then choose two (not necessarily distinct) elements to distinguish individually. In a less formal setting, the objects enumerated by this left side might be considered to be the number of ways to choose a k -person committee from n people, and then to select a chair and secretary (who might be the same person). In this context, the right side can be seen to enumerate the same thing through a casewise division: if the chair and secretary are the same, then we could preemptively choose the chair/secretary any of n ways, and then assign committee membership to the remaining n - 1 prospectives in any of 2 n - 1 ways. On the other hand, if the chair and secretary were different, we would choose any of the n people to be the chair, any of the remaining n - 1 to be the secretary, and then committ3ee membership for the remaining n - 2 people could be resolved in any of 2 n - 2 ways. There are other approaches possible, using either generating functions or direct algebra, but this is easily the simplest approach. 2. (12 of 12 students attempted this) Answer the following questions relating to paths through the following grid (note the excluded portions): (a) (5 points) How many walks are there from the lower left corner to the upper right corner taking upwards and rightwards steps only? If the grid were complete, there would be
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This note was uploaded on 01/12/2012 for the course MATH 681 taught by Professor Wildstrom during the Fall '09 term at University of Louisville.

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exam-1-091001-solutions - MATH 681 Exam #1 1. (8 of 12...

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