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

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

Hi JDPB7,

The subsets that include W are:



Final Answer:

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

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?

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)

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

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!

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.

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

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

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

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

What does the "2" represent in this case?

EMPOWERgmatRichC Bunuel ScottTargetTestPrep can you pls explain this 2^n formula/ concept? How has that come to being? And what does this signifiy

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