Possible Combinations with 2, 3, and 6: 3-Digit and 4-Digit Numbers

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For a 3-digit number using the digits 2, 3, and 6, there are 27 possible combinations since each digit can be repeated (3 choices for each of the 3 positions). In contrast, for a 4-digit number using the digits 1, 2, 3, and 4 without repetition, the total combinations are calculated as 4! (4 factorial), resulting in 24 unique arrangements. The discussion also clarifies that if there are 5 choices for a 3-digit number with repetition, the total is 125 (5^3), while without repetition, it would be 60 (5*4*3). The distinction between combinations and permutations is emphasized, particularly in relation to the second question. Understanding these principles is essential for solving similar problems effectively.
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Homework Statement


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.

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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?
 
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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?
 
helpingson said:
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.

You possibly meant "permutation", not combination. If you had meant the latter, the answer would be 1, not 4!.

helpingson said:
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?

Correct in all the cases. The last example is nothing but permutation of 5 things taken 3 at a time.
 
Thank You!
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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