The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x

A powerset of a set is the set of all subsets of that set. For example, the power set of {w, x, y} is {w}, {x}, {y}, {w, x}, {w, y}, {x,y}, {w, x, y}, and { } as given by the question stem. The cardinality of a powerset (the number of subsets of the powerset) is calculated by \(2^N\), where N is the number of elements in the set. Since {w, x, y} contains 3 elements, the cardinality of the powerset of that set contains \(2^3\) subsets.

The powerset of the set {w, x, y, z} contains \(2^4\) subsets. The powerset of the set {x, y, z} contains \(2^3\) subsets; these 8 subsets don't have w.
So the number of subsets of the set {w, x, y, z} that contain w is the total number of subsets minus number of subsets that don't contain w, 16-8 = 8.

Correct answer choice is D.

(Total number of sets) - (number of sets without w) = (number of sets with w)
2^4 - 2^3 = 8.
_________________

As an alternative approach, notice that the formula to get the the total number of subsets is

\(2^n\)
where n is the number of items in the set.

{w, x, y, z} has 4 in total so we have \(2^4=16\) subsets.
Since we want to see the number of subsets containing w, take it for granted that w is in your subset and use \(2^3\) instead.

EMPOWERgmatRichC, is there a way to answer this using combination formula?

Hi JDPB7,

The prompt itself literally tells you how to go about answering this question. You're asked for the total number of available subsets that contain W, and you're shown the 'definition' of what makes up a subset. With that knowledge, you should be able to list them all out (and there can't be that many, since the answers don't go any higher than 16).

If you try to create the list, then how many options do you come up with?

GMAT assassins aren't born, they're made,
Rich

Be careful with the question, because this a very tricky one!
Question asks us in how many sets is w present!
W is present in 1 {w} set
in 3 {w,...} set
and 3 {w,...,...} set
and one {w,x,y,z} set
total 8.

Answer is D

If you would like to count all combinations of the set {w,x,y,z}. Those are:
4+2C4+3C4+1+1=16 total (E)

Hi JDPB7,

The subsets that include W are:



Final Answer:

GMAT assassins aren't born, they're made,
Rich

Good official question as usual !

Just count number of subsets of {x,y,z} = 2^3 = 8, as we add w to each of these subsets, we get all subsets that contain w.

General formula
To get number of subsets including empty subset from a set of n numbers, nC0 + nC1 + .............. + nCn = 2^n

Set {w,x,y,z} = Total elements 4.

=> Therefore number of subsets : \(2^4\) =16.

Logic I:

=> Set without 'w' = { }, {x},{y},{z},{x,y},{x,z},{y,z},{x,y,z} = 8

Therefore, subsets of the set {w, x, y, z} contain w: 16 - 8 = 8


Logic II:

=> Set without 'w' means number of elements {x,y,z} = 3

Therefore number of subsets : \(2^3\) =8.

Therefore, subsets of the set {w, x, y, z} contain w: 16 - 8 = 8

Answer D

The long method to solve this question is as below:

Subset with 4 elements including w = 1
Subset with 3 elements including w = choosing 2 elements out of the remaining 3 = 3C2 = 3
Subset with 2 elements including w = choosing 1 element out of the remaining 3 = 3C1 = 3
Subset with just 1 element containing w = 1

Total Subsets =1 + 3 + 3 + 1 = 8

Answer D.



Thanks,
GyM
_________________

Certainly not the most efficient way, but worth giving a look:
The subsets containing w of the set {w , x , y , z}

{w}
{w , x}
{w , y}
{w , z}
{w , x , y}
{w , y , z}
{w , x , z}
{w , x , y , z}

Total 8 subsets.

Hi UNSTOPPABLE12,

Your list of the individual subsets assumes that at least one of the letters X, Y and Z exists in the set. That is NOT what the prompt states though - it asks how many subsets include "W." Those subsets would include all 7 of the ones that you listed as well as a set in which NONE of those 3 variables was included.

In reference to 2^N, you can think of each variable as either "in" or "not in", so there would be (2)(2)(2) = 8 possible sets.

GMAT assassins aren't born, they're made,
Rich

We can also solve this by Permutation & Combination method

We need W in _ or _ _ or _ _ _ or _ _ _ _

W in _ would be 1 way

W in _ _ would be 3C1 because W _(Any one of other 3 ) would be 3 ways

W in _ _ _ would be 3C2 because W _ _ (Any two of other 3 ) would be 3 ways

W in _ _ _ _ would be 1 way

Hence 1 + 3 + 3 +1 = 8

Hello woohoo921,


Yes, the formula used by Bunuel is a general formula.

Formula:

If a set S has ‘n’ elements, then the total number of subsets of S is 2n

Derivation:

• Let’s consider a random set S with n elements: {\(e_1\), \(e_2\), \(e_3\), ….. \(e_n\)}
• First observe that subsets are formed by using some elements and not using others from the main set.
  
  • For example: {1, 2} is a subset of {1, 2, 3} for which we kept both 1 and 2 in but 3 out. Similarly, {3} is a subset of {1, 2, 3} for which we kept only 3 in and kept both 1 and 2 out.
  • So, the main decision in creating any subset is on which elements to keep and which not to keep in the subset. In other words, for each element of the main set, we just need to make one out of two choices – keep this element (IN) OR do not keep this element (OUT).
• Coming back to set S, each element from all n has 2 choices: IN or OUT. We need to see the total number of ways in which we could take this decision for all n elements. That will give us the total number of subsets of S. (It now looks like a P&C question!)
  
  • Thus, total number of subsets would depend on the different ways n places can be filled with either IN or OUT.
  • The first place can be filled in 2 ways, second place can be filled in 2 ways, third place can be filled in 2 ways and so on and so forth.
  • Total number of subsets = 2*2*2*2*…. n times
  • = \(2^n\)

And that’s it!



Example:

Let’s take a very small set: {1, 2}.
We can quickly list and know that there are a total of 4 subsets – {1}, {2}, {1, 2}, {}.

Now, we’ll just verify that the formula gives us the same answer.
Since n = 2, total number of subsets = \(2^n\) = \(2^2\) = 4. That’s how simple the formula makes it!


Hope this helps!


Best Regards,
Ashish
Quant Expert, e-GMAT
_________________

I did it as a combinatoric.

Lets say we have 4 elements x, y, z and 0

we need to for all the posible arrangements for w _ _ _ , that means 4*3*2. The order doesnt matter then we divide 4*3*2/3= 8

Hi Elite097,

Based on how this prompt defines subsets, you can think of each of the letters (w, x, y and z) as either being "in" each subset or not - thus, that's 2 'options' for each letter. By extension, you can use multiplication (or raise 2 to a power) to find all of the possible subsets.

Mathematically - with the 4 letters given, that would be (2)(2)(2)(2) = 16 or 2^4 = 16

Listing them out, the 16 subsets would be

w
x
y
z
w and x
w and y
w and z
x and y
x and z
y and z
w, x and y
w, x and z
w, y and z
x, y and z
w, x, y and z
none of the four

As far as answering the specific question in the prompt, we're 'locking in' that the w will be in the subset, so that leaves just the 3 remaining letters (and whether they are in or not in the subset). That gives us (2)(2)(2) or 2^3 = 8 possible subsets that can be formed with the w included.

GMAT assassins aren't born, they're made,
Rich

Contact Rich at: Rich.C@empowergmat.com

Given: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x, y}, {w, x,y}, and { } (the empty subset).
Asked: How many subsets of the set {w, x, y, z} contain w ?

Subsets containing w = {w}, {w,x}, {w,y}, {w,z}, {w,x,y}, {w,y,z}, {w,z,x}, {w,x,y,z} : 8 subsets

IMO D

Is there a way to get to 8 from a combination method? I got 16 from finding the total combinations but how do I parse out which sets contain W and which ones do not? Just divide by 2 because the set has it or doesn't have it?

Obviously straight-listing works too!

chetan2u Gladiator59 Bunuel VeritasKarishma
Hello could anyone be kind to answer the following question?

So the general formula for finding subsets is 2^n, however when I try to separate the subsets of x,y,z manually I get 7 subsets rather than 8
(x),(y),(z) (x,y) (x,z) (y,z) (x,y,z) I don't get what is exactly the so called "empty subset" that is written in the stem I mean which combination is included in that subset...?

What does the "2" represent in this case?