Counting Digits

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/