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?
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.
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!".
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?
You possibly meant "permutation", not combination. If you had meant the latter, the answer would be 1, not 4!. Correct in all the cases. The last example is nothing but permutation of 5 things taken 3 at a time.