Last visit was: 17 Jul 2026, 17:10 It is currently 17 Jul 2026, 17:10
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
605-655 (Medium)|   Combinations|                              
User avatar
egmat
User avatar
e-GMAT Representative
Joined: 02 Nov 2011
Last visit: 17 Jul 2026
Posts: 5,956
Own Kudos:
33,665
 [3]
Given Kudos: 712
GMAT Date: 08-19-2020
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 5,956
Kudos: 33,665
 [3]
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 20 Jan 2023, 01:48
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
woohoo921
JDPB7
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). How many subsets of the set {w, x, y, z} contain w ?


(A) Four
(B) Five
(C) Seven
(D) Eight
(E) Sixteen

KarishmaB
If you have time, I would be so appreciative for your insights on this problem. Is the formula that Bunuel is using standardized and can be used for future problems? How do you come up with this formula? Thank you and Happy Holidays :)

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

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
User avatar
ahamd
User avatar
Joined: 25 Feb 2015
Last visit: 26 Jun 2024
Posts: 18
Own Kudos:
5
Given Kudos: 148
Posts: 18
Kudos: 5
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 31 Mar 2023, 06:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
JDPB7
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). How many subsets of the set {w, x, y, z} contain w ?


(A) Four
(B) Five
(C) Seven
(D) Eight
(E) Sixteen
Show more

1(selecting a single letter set)+3C1(Selecting two letters set->W is fixed & selecting the second letter from remaining 3 letters)+3C2(Selecting three letters set->W is fixed & selecting the other two letters from remaining 3 letters)+1(set with all four letters)

=8

Is it a right way to solve?
User avatar
tickledpink001
User avatar
Joined: 10 Dec 2021
Last visit: 28 Feb 2024
Posts: 31
Own Kudos:
425
Given Kudos: 4
Location: Australia
GMAT 1: 660 Q43 V47
GMAT 1: 660 Q43 V47
Posts: 31
Kudos: 425
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 30 Jan 2024, 03:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
thinkpad18
EMPOWERgmatRichC
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

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!

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

Thanks. Please can you help me understand how would we solve something like- a set contains 6 elements, how many ways exist to choose a subset of 3 from this set?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 17 Jul 2026
Posts: 112,012
Own Kudos:
828,108
Given Kudos: 110,130
Products:
GMAT Club Tests
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 112,012
Kudos: 828,108
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 30 Jan 2024, 03:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
tickledpink001

Thanks. Please can you help me understand how would we solve something like- a set contains 6 elements, how many ways exist to choose a subset of 3 from this set?
Show more

The number of ways to select a 3-element subset from a set containing 6 different elements is simply 6C3 = 20.
User avatar
pierjoejoe
User avatar
Joined: 30 Jul 2024
Last visit: 29 Jul 2025
Posts: 126
Own Kudos:
64
Given Kudos: 425
Location: Italy
Concentration: Accounting, Finance
GMAT Focus 1: 645 Q84 V84 DI78
GPA: 4
WE:Research (Technology)
GMAT Focus 1: 645 Q84 V84 DI78
Posts: 126
Kudos: 64
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 03 Sep 2024, 10:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­here the "event" that produces outcomes to count is the presence or absence from the set of a letter
in the example
x might be present or absent (2 outcomes)
y might be present or absent (2 outcomes)
w might be present or absent (2 outcomes)
total = 2^3

same applies for xyzw --> 2^4 = 16
now how many sets can be formed without w?
we take into account how many sets can be formed where we surely have w and than subtract this value from the total.

let's assume w is in every set--> we would have to arrange only the other 3 letters --> xyz. so how many way can we arrange them?
x might be present or absent (2 outcomes) (we assume each one have already w inside)
y might be present or absent (2 outcomes) (we assume each one have already w inside)
z might be present or absent (2 outcomes) (we assume each one have already w inside)
total = 2^3 = 8 possible outcomes where we have always w
(example --> if x, y, and z are absent the resulting set will be {w})
how many sets DO NOT contain w?

total = 16
sets containing w = 8

16 - 8 = 8 sets do not contain w
avatar
ManifestDreamMBA
avatar
Joined: 17 Sep 2024
Last visit: 01 Jul 2026
Posts: 1,383
Own Kudos:
915
Given Kudos: 243
Posts: 1,383
Kudos: 915
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 24 Apr 2025, 10:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
{w, x, y} has 2^3 subsets (including selecting none)

{w, x, y, z} will have 2^4 subsets.

Out of these half will have w, so subsets of the set {w, x, y, z} that contain w = 2^3 = 8
JDPB7
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). How many subsets of the set {w, x, y, z} contain w ?


(A) Four
(B) Five
(C) Seven
(D) Eight
(E) Sixteen
Show more
User avatar
luisdicampo
User avatar
Joined: 10 Feb 2025
Last visit: 29 Jun 2026
Posts: 479
Own Kudos:
91
Given Kudos: 328
Products:
My Reviews
Posts: 479
Kudos: 91
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 03 Apr 2026, 03:11
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Deconstructing the Question

We need to count how many subsets of \(\{w,x,y,z\}\) contain w.

The key idea is that w is required, so it is already fixed in every valid subset.

That means the only real choices are about x, y, and z.

Step-by-step

Start with w already included.

So every valid subset begins with:

\(\{w\}\)

Now decide what to do with x, y, and z.

For each of these 3 elements, there are 2 choices: include it or do not include it.

So the total number of valid subsets is:

\(2 \cdot 2 \cdot 2 = 2^3 = 8\)

To see this concretely, the valid subsets are:

\(\{w\}\)

\(\{w,x\}\)

\(\{w,y\}\)

\(\{w,z\}\)

\(\{w,x,y\}\)

\(\{w,x,z\}\)

\(\{w,y,z\}\)

\(\{w,x,y,z\}\)

So there are 8 subsets that contain w.

Answer: D
User avatar
PGTLrowanhand
User avatar
Joined: 30 Oct 2012
Last visit: 17 Jul 2026
Posts: 126
Own Kudos:
184
Given Kudos: 3
Status:London UK GMAT Consultant / Tutor
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 126
Kudos: 184
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 25 Jun 2026, 07:25
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Here's my video solution:
User avatar
anushree01
User avatar
Joined: 06 Apr 2024
Last visit: 17 Jul 2026
Posts: 232
Own Kudos:
80
Given Kudos: 170
Schools: HBS '26 ISB '26 INSEAD Jan '26
Products:
My Reviews
Schools: HBS '26 ISB '26 INSEAD Jan '26
Posts: 232
Kudos: 80
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, {x 25 Jun 2026, 19:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Simple and easy as always
Bunuel


(Total number of sets) - (number of sets without w) = (number of sets with w)
2^4 - 2^3 = 8.
Show more
   1   2 
Moderator:
Bunuel
Math Expert
112012 posts