Date: 10/21/98 at 17:42:59 From: Sharon Erle Subject: Combinatorics I came across this question in a text I'm using. It is supposed to be done with the Fundamental Principle of Counting. I know the answer and can generate it with a tree diagram or using combinations but not with the mn rule. How many six-digit numbers can you make with two zeros, two twos, and two fours? The solution is supposed to be 10(6)(1) and the section in the book that the problem is in assumes you don't know about factorials yet.
Date: 10/23/98 at 13:19:14 From: Doctor Rob Subject: Re: Combinatorics Hi Sharon, here one approach. How many ways are there to pick the two spots for the zeroes from the six digit locations? Answer: 5*4/2 = 10 (you can't put a zero at the beginning). Once you have placed the two zeroes, how many ways are there to pick the two spots for the twos from the four remaining digit locations? Answer: 4*3/2 = 6. Once you have placed the zeroes and twos, how many ways are there to pick the two spots for the fours from the two remaining digit locations? Answer: 1, because you have no freedom of choice left. The final answer is then 10*6*1 = 60. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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