Good point!
The question assumes that all elements are different.
I'll edit it accordingly.

I have a query here in option A we ask subsets with odd number of elements.
And in question it mentions 35 different elements, how will we know that odd and even numbers are in equal proportion?

I approached this in a different way, please correct me if I thought about this the wrong way GreenlightTestPrep

Essentially, it doesn't matter what the elements are. They can be symbols representing animals for example. The important thing is that they are different. To keep it simple then, let's imagine the set of 35 different elements are the letters a-z (26 letters) and 9 capital letters (A-I), making a set of 35 different elements.

Since there are 35 elements, let's create subsets with an odd number of elements in them. How many are there?

1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35.

There are 18 different subsets we can make, each one containing an odd number of elements (one subset as one element, another has three elements, another has 5 elements, etc..)

So now lets take these 35 different elements:

a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z,A,B,C,D,E,F,G,H,I

And we're going to put them into each odd-numbered subset.

How many ways can you put these 35 different elements into a subset that can hold only 1 element? \(35C1\)
How many ways can you put these 35 different elements into a subset that can hold only 3 elements? \(35C3\)
How many ways can you put these 35 different elements into a subset that can hold only 5 elements? \(35C5\)
How many ways can you put these 35 different elements into a subset that can hold only 7 elements? \(35C7\)
.
.
.
How many ways can you put these 35 different elements into a subset that can hold only 33 elements? \(35C33\)
How many ways can you put these 35 different elements into a subset that can hold only 35 elements? \(35C35\)

So the total amount of odd-numbered sets can be written out in the following way:

\(35C1\) + \(35C3\) + \(35C5\) + \(35C7\) + .... + \(35C33\) + \(35C35\)

This number is enormous and isn't worth going through the trouble to calculate fully. But we can calculate \(2^{18}\) pretty simply with a calculator.

Note that \(2^{10}\) = \(1,024\) and \(2^8\) = \(2^{10}\) / \(2^2\) = \(1,024\) / \(4\) = \(256\).

So \(2^{18}\) = \(1,024 * 256\) = \(262,144\)

Now let's just try evaluating the first couple of combination formulas from above and see what happens (again, using a calculator):

\(35C1\) = \(35\)

\(35C3\) = \( 6,545 \)

\(35C5\) = \(324,632\)

At this point, we can stop here. The sum of the first three numbers is already bigger than \(2^{18}\). It must be then that the whole sum is greater than \(2^{18}\).

Therefore the answer is A.

GreenlightTestPrep , I agree with grenico , the prompt seems to be asking about the number of different odd-sized subsets (i.e. number of unique 1-element subsets + number of unique 3-element subsets + number of unique 5-element subsets + ... + number of unique 35-element subsets), not the number of unique subsets containing only "odd-indexed" elements:








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(I don't have an extensive enough post history to directly link the URLs here)

-> Binomial Coefficient:

The binomial coefficient C(n, k) is the number of ways of picking k unordered outcomes from n possibilities, also known as a combination or combinatorial number.

...

C(n, k) therefore gives the number of k-subsets possible out of a set of n distinct items. For example, The 2-subsets of {1,2,3,4} are the six pairs {1,2}, {1,3}, {1,4}, {2,3}, {2,4}, and {3,4}, so C (4, 2) = 6.

(Weisstein, Eric W. "Binomial Coefficient." From MathWorld--A Wolfram Web Resource. search "wolfram Binomial Coefficient")


-> k-Subset:

A k-subset is a subset of a set on n elements containing exactly k elements. The number of k-subsets on n elements is therefore given by the binomial coefficient C(n, k). For example, there are C (3, 2) = 3 2-subsets of {1,2,3}, namely {1,2}, {1,3}, and {2,3}.

(Weisstein, Eric W. "k-Subset." From MathWorld--A Wolfram Web Resource. search "wolfram k-Subset")

The total number of distinct k-subsets on a set of n elements (i.e., the number of subsets) is given by
sum_(k=0)^n C(n, k)=2^n.

---

-> Count of Subsets with Even Cardinality:

Theorem

Let S
be a set whose cardinality is n

Then the number of subsets of S
whose cardinality is even is 2^(n−1)

search "proofwiki Count of Subsets with Even Cardinality"


-> Count of Subsets with Odd Cardinality:

Theorem

Let S
be a set whose cardinality is n

Then the number of subsets of S
whose cardinality is odd is 2^(n−1)

search "proofwiki Count of Subsets with Odd Cardinality"

-> Symmetry Rule for Binomial Coefficients:

Theorem

Let n∈Z>0,k∈Z

Then:

C(n, k) = C(n, n−k)

search "proofwiki Symmetry Rule for Binomial Coefficients"



----
So for this prompt...

B: 2^18
A:

For every k∈Z∈[0, 35], there are a total of sum(C(35,k))= sum(C(35, k_odd)) + sum(C(35,k_even)) = 2^35 different unique size-k subsets of a set composed of 35 different elements.

C(35,0) + C(35,1) + C(35,2) + C(35,3) + C(35,4) + C(35,5) + C(35, 6) + C(35,7) + C(35,8) + C(35,9) + C(35,10) + C(35,11) + C(35,12) + C(35,13) + C(35,14) + C(35,15) + C(35, 16) + C(35,17) + C(35,18) + C(35,19) + C(35,20)+ + C(35,21) +C(35,22) + C(35,23) + C(35,24) + C(35,25) + C(35, 26) + C(35,27) + C(35,28) + C(35,29) + C(35,30) + C(35,31) + C(35,32) + C(35,33) + C(35,34) + C(35,35) = 2^35

With n = 35 , there should be 2^(n-1) = 2^34 k_odd-sized subsets, and 2^(n-1) = 2^34 k_even-sized subsets...

The symmetry rule for binomial coefficients shows that:

C(35,0) = C(35,35)
C(35,1) = C(35,34)
C(35,2) = C(35,33)
C(35,3) = C(35,32)
C(35,4) = C(35,31)
C(35,5) = C(35,30)
C(35,6) = C(35,29)
C(35,7) = C(35,28)
C(35,8) = C(35,27)
C(35,9) = C(35,26)
C(35,10) = C(35,25)
C(35,11) = C(35,24)
C(35,12) = C(35,23)
C(35,13) = C(35,22)
C(35,14) = C(35,21)
C(35,15) = C(35,20)
C(35,16) = C(35,19)
C(35,17) = C(35,18)

There are 18 terms in each column, and each row has an odd-even pair...for each quantity of unique k_odd-sized subsets, there is matching quantity of unique k_even-sized subsets.

So the number of unique odd-sized subsets = number of even-sized subsets for a set with n = 35 different elements:
= (2^35)/2
= 2^(35-1)
= 2^34




--------------------------
A = 2^34
B = 2^18

A > B

so the answer is A

I'm not sure if a data analysis/combinatorial question of this nature would be on the exam, however, level 5 problems might require similar logical deductions.

In other words, the advantage of knowing the theorems is small, but the logical deductions that I and GreenLight used to arrive at the answer are way more significant. You might have to go through similar mental hoola-hoops in word problems, geometry, or algebra that make use of simpler principles.

So don't focus on the answer, but the process of problem-solving itself. That is what is needed to excel in the GRE math. Hope this helps.

The number of subsets of A containing odd number of elements

Take the task of creating subsets and break it into stages.
Let's label for 35 elements as element1, element2, element3,......element35

Stage 1: Determine whether to include element1 in the subset
We can choose to include element1 in the subset, OR we can choose to NOT include element1 in the subset
So, we can complete stage 1 in 2 ways

Stage 2: Determine whether to include element2 in the subset
Applying the same logic, we can complete this stage in 2 ways

Stage 3: Determine whether to include element3 in the subset
We can complete this stage in 2 ways

.
.
.
Stage 35: Determine whether to include element35 in the subset
We can complete this stage in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 35 stages (and thus create a subset) in (2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= \(2^{35}\) )

IMPORTANT: We can create \(2^{35}\) different subsets from the 35 elements.
However, the question asks us to determine the number of subsets that contain an ODD number of elements
The key here is to recognize that among the many possible subsets, HALF will contain an ODD number of elements, and HALF will contain an EVEN number of elements.

So, will take the total number of possible subsets, \(2^{35}\), and divided by 2 to get \(\frac{2^{35}}{2}\)
\(\frac{2^{35}}{2}=\frac{2^{35}}{2^1}= 2^{34}\)

We get:
QUANTITY A: \(2^{34}\)
QUANTITY B: \(2^{18}\)

Answer: A

Note: the FCP can be used to solve the MAJORITY of counting questions on the GRE. So, be sure to learn it.



Sir, how can we assume there are half even and half odd number? what if all the 35 numbers are even?

Quantity A has nothing to do with whether the elements are odd or even. In fact, the 35 elements could all be letters, or algebraic expressions, or anything else.

Instead, quantity A refers to the number of elements in a subset.
For example, if set A consists of the names of 35 countries, then the subset {Canada, India} has an odd number of elements.

Does that help?

thank you sir, my bad, I miss read the question.

The question is referring to number of sets and not numbers in the set!

Based on 35 different elements and The number of subsets of A containing odd number of elements.
It can be in many ways.
Case 1,
none of 35 different elements are odd. For example, all of them are even like 2,4,6,8........
Case 2,
all of them are odd. For example, 1,3,5,7.........

Please guide me I am so confused with this one.
Is it an official question?

The explanation provided by Brent above https://gre.myprepclub.com/forum/sets-w ... tml#p42751 is robust.

You should refer to it Sir

The question is about the odd number of subsets of the set and not number of subsets containing odd numbers.

There is direct formula for combination when we have distinct items in the set and we need select the number of items.

suppose we have A B C D E then we can select it like 5C0 + 5C1 + 5C2 +5C3+ 5C4 + 5C5 which equal to 2^n (n = number of distinct items)

thus for 35 elements we we can do selection in 2^35 ways but we want selection of only odd elements , so as we know we have equal number
of odd and even elements from 1 to 35 thus we can divide number of ways by 2.

2^35/2 = 2^34