Algebra 2 help please??

Question

Consider the set x= { #, $, @, ^, *}
1. Find the numbers of subsets there are for set X.
2. Find the number of 4- element subset of set X.

I am so confused, anything helps, thank you!!!

Answer 1

Hi! (I'm going to be honest with you, this doesn't really look like an Algebra 2 problem to me, but okay...)

 

First of all, a set is a grouping of elements that can be anything, where each element only appears ONCE. You can have an empty set (aka, the null set), {}; you can a set of words, or numbers, or both; you can have a set of letters, other characters, strings of non-word characters; you can even have sets of other sets! Basically, a set is a list of anything under the sun, and they don't really have to have anything in common except that they are all in your set.  In your case, the set contains the elements #, $, @, ^, *.

 

Now, a subset is a set where all of the elements are contained in another set.  Two important things to note is that the empty set (aka the null set), is a subset of everything, and that all sets, including the null set, are subsets of themselves.  For example, if I have a set A which I define as A={a, b, c}, and a set B where B= {b,c}, B is a subset of A because A has all the elements of B. Have I lost you yet? I hope not!

 

Next, we look at your set in particular.  Because we know that all sets are subsets of themselves, and that the null set is a subset of every set, we already have 2 subsets found! Woo-hoo!

So, how do we find the rest? Simple! We use the _nC_r equations from probability.

There are ₅C₁ single element subsets: 5!/(1!*4!) = 5*4*3*2*1/(1*4*3*2*1) = 5

There are ₅C₂ single element subsets: 5!/(2!*3!) = 5*4*3*2*1/(2*1*3*2*1) = 5*2 = 10

There are ₅C₃ single element subsets: 5!/(3!*2!) = 5*4*3*2*1/(3*2*1*2*1) = 5*2 = 10

There are ₅C₄ single element subsets: 5!/(4!*1!) = 5*4*3*2*1/(4*3*2*1*1) = 5

 

So, we have the two 'givens', and then 5, 10, 10, and 5. 2+5+10+10+5 = 32 total subsets. When we were calculating, we showed that there were 5 4 element subsets.

 

(A quick way to find the total number is to calculate 2ⁿ where n is the number of elements in your set.  This finds the number of elements of the power set, which is the set of all subsets of a given set. Here, there are 5 elements, so we would calculate 2⁵=32.)

Answer 2

You've got a set with 5 elements. Let's figure out how many 0-element subsets there are, how many 1-element subsets there are, how many 2-element subsets there are, etc.

 

 

 

0-element subsets:

 

There is only 1, the null set. {ø}

 

 

1-element subsets:

 

There are 5 of these, one for each element.

 

 

2-element subsets:

 

There are (5 choose 2) of these, for a total of 10.

 

(In case you don't remember combinations, that's ₅C₂ = 5!/(2!(5-2)!) = (5×4×3×2×1)/(2×1(3!)) = (5×4×3)/(3×2×1) = (5×4)/2 = 20/2 = 10.)

 

 

3-element subsets:

 

There are (5 choose 3) of these, for a total of 10.

 

 

4-element subsets:

 

There are (5 choose 4) of these, for a total of 5.

 

 

5-element subsets:

 

There is only 1 of these. {#, $, @, ^, *}

 

 

That's a total of 1 + 5 + 10 + 10 + 5 + 1 = 32 subsets. Of those 32 subsets, there are 5 4-element subsets.