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

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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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New post 06 Nov 2015, 13:28
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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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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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New post 07 Nov 2015, 15:10
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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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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y},  [#permalink]

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New post 06 Nov 2015, 15:41
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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?

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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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New post 07 Nov 2015, 14:11
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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)


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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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New post 25 Dec 2015, 17:17
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?
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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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New post 26 Dec 2015, 05:55
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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New post 25 Dec 2016, 00:13
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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New post 23 Mar 2017, 04:35
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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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New post 03 Jun 2017, 02:29
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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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New post 02 May 2018, 12:22
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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New post 09 May 2018, 13:58
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!
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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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New post 09 May 2018, 21:42
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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New post 20 Sep 2018, 13:12
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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\(2^4 - 2^3\)

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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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New post 20 Sep 2018, 20:54
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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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New post 21 Sep 2018, 23:17
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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 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.



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Re: The subsets of the set {w, x, y} are {w}, {x}, {y}, {w, x}, {w, y}, &nbs [#permalink] 21 Sep 2018, 23:17
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