Unformatted text preview: To count the number of solutions to this equation consider the number of equations with x i ≥ 1 and ∑ n i =1 x i = ˆ k , which is ˆ k 1 n 1 so to calculate the number of equations to the requested problem we add these up for all ˆ k < k . The number of solutions is given by k ˆ k = n ˆ k 1 n 1 with k > n. Chapter 1: SelfTest Problems and Exercises Problem 1 (counting arrangements of letters) Part (a): Consider the pair of A with B as one object. Now there are two orderings of this “fused” object i.e. AB and BA . The remaining letters can be placed in 4! orderings and once an ordering is specified the fused A/B block can be in any of the five locations around the permutation of the letters CDEF . Thus we have 2 · 4! · 5 = 240 total orderings. Part (b): We want to enforce that A must be before B . Lets begin to construct a valid sequence of characters by first placing the other letters CDEF , which can be done in 4! = 24 possible ways. Now consider an arbitrary permutation ofpossible ways....
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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.
 Winter '08
 G.JOGESHBABU
 Probability

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