P0028_Soln - 9-4-3 2 = 9 4 5 3 2 2 = 9 4 5 3

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P0028_Soln.doc In other words, how many ways can we randomly partition 9 positions into 3 groups with the size of each group being the number of units of each type ( e.g., 4 units type A)? 9! 4!3!2! = 9 4 9 - 4 3
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Unformatted text preview: 9-4-3 2 = 9 4 5 3 2 2 = 9 4 5 3...
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This note was uploaded on 02/01/2011 for the course BME 335 taught by Professor Dunn during the Spring '10 term at University of Texas at Austin.

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