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Intern  Joined: 07 Mar 2015
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The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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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

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Manager  Joined: 01 Jan 2015
Posts: 62
Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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JDPB7 wrote:
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

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.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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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,
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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ EMPOWERgmat Instructor V Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 14823 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, [#permalink] ### Show Tags 2 1 Hi JDPB7, The subsets that include W are: {W}, {W, X}, {W, Y}, {W, Z}, {W, X, Y}, {W, X, Z}, {W, Y, Z} and {W, X, Y, Z) Final Answer: GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com *****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!***** # Rich Cohen Co-Founder & GMAT Assassin Follow Special Offer: Save$75 + GMAT Club Tests Free
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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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?
Intern  Joined: 31 Aug 2013
Posts: 9
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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JDPB7 wrote:
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

-------------------------
The number of ways of selecting 0 or more from n elements is 2^n.
Hence here, once you select w , the number of ways of selecting 0 or more from x,y and z ( 3 elements) is 2^3 = 8.

The same applies for {w, x, y}. If you select w here, the number of ways of selecting 0 or more from x and y = 2^2 = 4 (which aligns with the actual results)

Hope it helps.
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
Total number of subset with 03 elements : 08 (given in the problem)
Total number of subset with 04 elements ( with one element repeated) = Total number of subset with 03 elements ( As the 4th element is forced in all subsets)

Hence Answer = 08 (D)
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
1
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)
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
its just a counting problem

such as how many menus are possible

thus, (w,x,y) --> 2*2*2=8
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
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
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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EMPOWERgmatRichC wrote:
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!
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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thinkpad18 wrote:
EMPOWERgmatRichC wrote:
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.

Explained here: https://gmatclub.com/forum/the-subsets- ... l#p1598828
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
Bunuel wrote:
thinkpad18 wrote:
EMPOWERgmatRichC wrote:
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 - 2^2 = 8.

Explained here: https://gmatclub.com/forum/the-subsets- ... l#p1598828

Hi Bunuel,
I guess the highlighted portion is a typo. should it be$$2^4 - 2^3$$

Probus
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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Probus wrote:
Hi Bunuel,
I guess the highlighted portion is a typo. should it be$$2^4 - 2^3$$

Probus

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Yes. Edited. Thank you.
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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1
JDPB7 wrote:
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

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
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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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JDPB7 wrote:
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

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.
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The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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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...?
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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2
UNSTOPPABLE12 wrote:
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...?

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
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GMAT 1: 740 Q50 V40 GMAT 2: 770 Q51 V42 Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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Hi UNSTOPPABLE12, Welcome to GMAT Club!

You are absolutely right. There are $$2^n$$ subsets of a set with a cardinality of n.
So for a 3 element set {x,y,z} the subsets will be..
{x},{y},{z},{x,y},{x,z},{y,z},{x,y,z} and the empty set denoted by {}

Note that an empty set is a subset of all sets.

Hope this answers your question. Feel free to tag back if you have additional queries.

Best,
Gladi

UNSTOPPABLE12 wrote:
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...?

_________________
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Gladi

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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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Gladiator59 wrote:
Hi UNSTOPPABLE12, Welcome to GMAT Club!

You are absolutely right. There are $$2^n$$ subsets of a set with a cardinality of n.
So for a 3 element set {x,y,z} the subsets will be..
{x},{y},{z},{x,y},{x,z},{y,z},{x,y,z} and the empty set denoted by {}

Note that an empty set is a subset of all sets.

Hope this answers your question. Feel free to tag back if you have additional queries.

Best,
Gladi

UNSTOPPABLE12 wrote:
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...?

Gladiator59 Thank you for your prompt response ! Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},   [#permalink] 02 Jan 2019, 05:11

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