
#1
Apr208, 06:36 PM

P: 3

1. The problem statement, all variables and given/known data
Q #1  A math teacher wants to give each student a 3 digit number using only the numbers 2, 3 and 6. Numbers can be repeated. How many possible combinations are there? Q #2  (simplified) How many possible four digit combinations are there for the numbers 1, 2, 3 and 4 only using each number once in each combination. 2. Relevant equations 3. The attempt at a solution A #1  I come up with 27, but I know there is a formula that will help me reach that number without writing each possibility out. A #2  I remember from years ago in my HS days a formula that was something like: 4 x 3 x 2 x 1 to figure out this type of questions, but maybe I am way off. Any ideas? 



#2
Apr208, 08:29 PM

PF Gold
P: 1,132

How many ways can the first number be selected for the 3 digit number? There are 3 choices right? Now consider the number of ways which the second number can be selected. Since they can be repeated, there are 3 possibilities again. At this stage, there are 3 starting numbers, and each starting number is followed by 1 of 3 other numbers, giving a total of 3*3=9 combinations. This is only for combinations of 2 different numbers, so now try and apply the same theory to 3.




#3
Apr308, 05:59 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,900

For number 2 your answer is correct. You have 4 choices for the first number. Once you have chosen that, you can't use it again so you have 3 choices for the second number. Now you can't use either of the first two numbers so you have 2 choices for the third number. Of course, there is only 1 number left for the fourth. The total number of ways you could choose is the product of all those: 4*3*2*1, also known as "4!".




#4
Apr308, 10:57 AM

P: 3

Possible combinations
Thank you for the replys!!! I want to make sure I understand #1: If we were dealing with a 3 digit number, but had 5 choices (1, 2, 3, 4, 5), would there be 125 possibilities? 5*5*5=125. To complicate things, what if we were dealing with a 3 digit number, had 5 choices, but the numbers could not be repeated? Would it be 5*4*3=60?




#5
Apr508, 06:20 PM

HW Helper
P: 1,986





#6
Apr608, 07:09 PM

P: 3

Thank You!



Register to reply 
Related Discussions  
combinations ice cream math  Calculus & Beyond Homework  6  
How Many Different Combinations?  Precalculus Mathematics Homework  11  
Sum of Combinations  Set Theory, Logic, Probability, Statistics  12  
four tires to mount on a car these can be mounted  Set Theory, Logic, Probability, Statistics  3  
Combinations  Introductory Physics Homework  4 