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Unformatted text preview: The number of solutions to such integer equalities is discussed in the book. Specifically, there are 3 + 4 1 4 1 = 6 3 = 20 , such solutions. For each of these solutions, we have 3! = 6 ways to place the three other letters giving a total of 6 · 20 = 120 arraignments. Part (d): For this problem A must be before B and C must be before D . Let begin to construct a valid ordering by placing the letters E and F first. This can be done in two ways EF or FE . Next lets place the letters A and B , which if A is located at the left most position as in AEF , then B has three possible choices. As in Part (b) from this problem there are a total of 3 + 2 + 1 = 6 ways to place A and B such that A comes before B . Following the same logic as in Part (b) above when we place C and D there are 5 + 4 + 3 + 2 + 1 = 15 possible placements. In total then we have 15 · 6 · 2 = 180 possible orderings. Part (e): There are 2! ways of arranging A and B , 2! ways of arranging C and D , and 2!...
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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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