Solutions from a student (dragged) 15

# Solutions from a student (dragged) 15 - choice v 1 = x 1 2...

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Unformatted text preview: choice v 1 = x 1- 2 ≥ v 2 = x 2- 2 ≥ v 3 = x 3- 3 ≥ v 4 = x 4- 4 ≥ 1 1 1 1 2 1 1 1 3 1 1 1 4 1 1 1 Table 1: All possible choices of three investments. 11, and has 11 + 3- 1 3- 1 = 13 2 = 78 possible solutions. Investment choice number two requires the equation v 1 + v 3 + v 4 = 20- 2- 3- 4 = 11, and again has 11 + 3- 1 3- 1 = 13 2 = 78 possible solutions. Investment choice number three requires the equation v 1 + v 2 + v 4 = 20- 2- 2- 4 = 12, and has 12 + 3- 1 3- 1 = 14 2 = 91 possible solutions. Finally, investment choice number four requires the equation v 1 + v 2 + v 3 = 20- 2- 2- 3 = 13, and has 13 + 3- 1 3- 1 = 15 2 = 105 possible solutions. Of course we could also invest in all four opportunities which has the same number of possibilities as in part (a) or 220. Then in total since we can do any of these choices we have 220 + 105 + 91 + 78 + 78 = 572 choices. Chapter 1: Theoretical Exercises Problem 1 (the generalized counting principle) This can be proved by recursively applying the basic principle of counting.This can be proved by recursively applying the basic principle of counting....
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## This note was uploaded on 02/25/2011 for the course STAT 418 taught by Professor G.jogeshbabu during the Winter '08 term at Penn State.

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