How Do You Enumerate All Subsets of a Given Set?

AI Thread Summary
To enumerate all subsets of the set B = {4, 8, 12}, the formula 2^n is used, where n is the number of elements in the set. For set B, n equals 3, resulting in 2^3 = 8 total subsets. The subsets include the empty set, individual elements, pairs of elements, and the full set itself. Understanding the structure of subsets involves recognizing that they can be categorized by the number of elements they contain. This method provides a systematic way to list all possible subsets.
Craß
Messages
3
Reaction score
0

Homework Statement


List all the subsets of set B.
B = {4,8,12}


Homework Equations


A={5,10,15,20} C={4,8,12,16} D={2,4,6,8,10} E={4,12}


The Attempt at a Solution


I know that the equation for finding the number of subsets is 2n, but I don't exactly understand how I'm supposed to list them.
 
Physics news on Phys.org
Craß said:

Homework Statement


List all the subsets of set B.
B = {4,8,12}


Homework Equations


A={5,10,15,20} C={4,8,12,16} D={2,4,6,8,10} E={4,12}
How are these relevant to your problem?
Craß said:

The Attempt at a Solution


I know that the equation for finding the number of subsets is 2n, but I don't exactly understand how I'm supposed to list them.
First off, 2n is not an equation. To use this expression you need to know what n is in your problem, which is the number of elements in the set.
 
Could you please at least answer my question instead of pointing out the flaws in my post? I think that it's pretty straight forward even though I don't know what the hell I'm talking about. But hey, that's why I'm asking for help.
 
What is n in your problem? When you answer that, you'll have a start on figuring things out.
 
In general, a set with n elements has the following subsets:
The empty set.
The one-element subsets.
The two-element subsets.
...
The subset with n elements (the set itself).
 
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
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
Back
Top