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Sets with odd number of elements [#permalink]
27 Nov 2019, 19:38

00:00

Question Stats:

26% (01:07) correct
73% (00:22) wrong based on 15 sessions

Set A contains 35 elements.

Quantity A

Quantity B

The number of subsets of A containing odd number of elements

\(2^{18}\)

A)The quantity in Column A is greater. B)The quantity in Column B is greater. C)The two quantities are equal. D)The relationship cannot be determined from the information given.

Re: Sets with odd number of elements [#permalink]
04 Dec 2019, 11:29

2

This post received KUDOS

Expert's post

novice07 wrote:

Set A contains 35 elements.

Quantity A

Quantity B

The number of subsets of A containing odd number of elements

\(2^{18}\)

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.

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Re: Sets with odd number of elements
[#permalink]
04 Dec 2019, 11:29