Q&A #19440

Teachers' Lounge Discussion: Combinations

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From: Teh Oh Kian

To: Teacher2Teacher Public Discussion
Date: 2010111705:34:06
Subject: Combinations

Dear both, I think we should do in this way: The first committee member can be chosen in any of the 5 persons, second from any of the 4 and third from any of the 3, but since order of choice is not important, and so repeated cases (3x2x1 ways)must be omitted. As such the total number of ways = (5x4x2)/(3x2x1) = 10 Regards Teh

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