GRE Math : Arithmetic

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Example Questions

Example Question #2 : Combinations

There are eight possible flavors of curry at a particular restaurant.

Quantity A: Number of possible combinations if four unique curries are selected.

Quantity B: Number of possible combinations if five unique curries are selected.

Possible Answers:

The two quantities are equal

Quantity A is greater.

The relationship cannot be determined.

Quantity B is greater.

Correct answer:

Quantity A is greater.

Explanation:

The number of potential combinations for selections made from possible options is

$C(n,k)=\frac{n!}{(n-k)!k!}$

Quantity A:

$C(8,4)=\frac{8!}{(8-4)!4!}=70$

Quantity B:

$C(8,5)=\frac{8!}{(8-5)!5!}=56$

Quantity A is greater.

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Example Question #11 : How To Find The Greatest Or Least Number Of Combinations

Quantity A: The number of possible combinations if four unique choices are made from ten possible options.

Quantity B: The number of possible permutations if two unique choices are made from ten possible options.

Possible Answers:

The relationship cannot be established.

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

Correct answer:

Quantity A is greater.

Explanation:

For choices made from possible options, the number of potential combinations (order does not matter) is

$C(n,k)=\frac{n!}{(n-k)!k!}$

And the number of potential permutations (order matters) is

$P(n,k)=\frac{n!}{(n-k)!}$

Quantity A:

$C(10,4)=\frac{10!}{(10-4)!4!}=210$

Quantity B:

$P(10,2)=\frac{10!}{(10-2)!}=90$

Quantity A is greater.

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Example Question #3 : Combinations

There are possible flavor options at an ice cream shop.

$\textup{Quantity A: The number of possible combinations if three unique flavors are selected.}$

$\textup{Quantity B: The number of possible combinations if five unique flavors are selected.}$

Possible Answers:

$\textup{The two quantities are equal}$

$\textup{Quantity B is greater}$

$\textup{Quantity A is greater}$

$\textup{The relationship cannot be determined}$

$\textup{One of the numbers is not a real number}$

Correct answer:

$\textup{Quantity B is greater}$

Explanation:

When dealing with combinations, the number of possible combinations when selecting choices out of options is:

$C(n,k)=\frac{n!}{(n-k)! k!}$

For Quantity A, the number of combinations is:

$C(10,3)=\frac{10!}{(10-3)!3!}$

C(10,3)=120

For Quantity B, the number of combinations is:

$C(10,5)=\frac{10!}{(10-5)!5!}$

C(10,5)=252

Quantity B is greater.

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Example Question #3 : Combinations

Quantity A: The number of potential combinations given two choices made from ten options.

Quantity B: The number of potential combinations given four choices made from twenty options.

Possible Answers:

The relationship cannot be determined.

Quantity B is larger.

Quantity A is larger.

The two quantities are equal.

Correct answer:

Quantity B is larger.

Explanation:

Since in this problem we're dealing with combinations, the order of selection does not matter.

With selections made from potential options, the total number of possible combinations is

$C(n,k)=\frac{n!}{(n-k)!k!}$

Quantity A:

$C(10,2)=\frac{10!}{(10-2)!2!}=45$

Quantity B:

$C(20,4)=\frac{20!}{(20-4)!4!}=4845$

Quantity B is larger.

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Example Question #331 : Arithmetic

Quantity A: The total number of combinations when a combination of two choices is made out of ten options and a combination of one choice is made from two options.

Quantity B: The total number of permutations formed when two choices are made from ten options.

Possible Answers:

Quantity B is greater.

The two quantities are equal.

The relationship cannot be established.

Quantity A is greater.

Correct answer:

The two quantities are equal.

Explanation:

With selections made from potential options, the total number of possible combinations (order doesn't matter) is:

$C(n,k)=\frac{n!}{(n-k)!k!}$

With selections made from potential options, the total number of possible permutations(order matters) is:

$P(n,k)=\frac{n!}{(n-k)!}$

Quantity A:

This quantity involves two combinations being made to create one grand combination, the number of potential results of which is equal to the product of the subresults.

$C(10,2)=\frac{10!}{(10-2)!2!}=45$

$C(2,1)=\frac{2!}{(2-1)!1!}=2$

The total number of combinations is $2 \cdot 45 = 90$

Quantity B:

$P(10,2)=\frac{10!}{(10-2)!}$

The total number of permuations is

The two quantities are equal.

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Example Question #4 : Combinations

Quantity A: The number of combinations if five choices are made from ten options.

Quantity B: The number of combinations if two choices are made from twenty options.

Possible Answers:

The relationship cannot be determined.

Quantity A is greater.

The two quantities are equal.

Quantity B is greater.

Correct answer:

Quantity A is greater.

Explanation:

Since we're dealing with combinations in this problem, the order of selection does not matter.

With selections made from potential options, the total number of possible combinations is

$C(n,k)=\frac{n!}{(n-k)!k!}$

Quantity A:

$C(10,5)=\frac{10!}{(10-5)!5!}=252$

Quantity B:

$C(20,2)=\frac{20!}{(20-2)!2!}=190$

Quantity A is greater.

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Example Question #24 : How To Find The Greatest Or Least Number Of Combinations

Davis is trying to decide on an ensemble. He has ten pairs of socks to choose from, six bracelets (he'll wear a different one on each wrist), and four pairs of sun glasses. How many ways can he create an outfit?

Possible Answers:

1200

304

600

Correct answer:

1200

Explanation:

With selections made from potential options, the total number of possible combinations (order doesn't matter) is:

$C(n,k)=\frac{n!}{(n-k)!k!}$

With selections made from potential options, the total number of possible permutations(order matters) is:

$P(n,k)=\frac{n!}{(n-k)!}$

When choosing the socks and sunglasses, order does not matter; however, in electing bracelets, order does matter: for instance, a blue on the left and green on the right is different than a green on the left and blue on the right!

Consider the results case by case:

Socks:

$C(10,1)=\frac{10!}{(10-1)!1!}=10$

Bracelets:

$P(6,2)=\frac{6!}{(6-2)!}=30$

Sunglasses:

$C(4,1)=\frac{4!}{(4-1)!1!}=4$

The number of potential ensembles is the product of these:

10(30)(4)

1200

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Example Question #83 : Probability & Statistics

Lisa is dressing warm for the winter. She'll be layering three shirts over each other, and two pairs of socks. If she has fifteen shirts to choose from, along with ten different kinds of socks, how many ways can she layer up?

Possible Answers:

20475

2820

245700

500

Correct answer:

245700

Explanation:

Since the order in which Lisa layers up matters, we're dealing with permutations.

With selections made from potential options, the total number of possible permutations is:

$P(n,k)=\frac{n!}{(n-k)!}$

For her shirts:

$P(15,3)=\frac{15!}{(15-3)!}=2730$

For her socks:

$P(10,2)=\frac{10!}{(10-2)!}=90$

Her total ensemble options is the product of these two results
90(2730)

245700

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Example Question #331 : Arithmetic

Quantity A: The number of possible combinations when three choices are made from six options.

Quantity B: The number of possible permutations when three choices are made from four options.

Possible Answers:

The two quantities are equal.

The relationship cannot be determined.

Quantity B is greater.

Quantity A is greater.

Correct answer:

Quantity B is greater.

Explanation:

With selections made from potential options, the total number of possible combinations (order doesn't matter) is:

$C(n,k)=\frac{n!}{(n-k)!k!}$

With selections made from potential options, the total number of possible permutations(order matters) is:

$P(n,k)=\frac{n!}{(n-k)!}$

Quantity A:

$C(6,3)=\frac{6!}{(6-3)!3!}=20$

Quantity B:

$P(4,3)=\frac{4!}{(4-3)!}=24$

Quantity B is greater.

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Example Question #21 : How To Find The Greatest Or Least Number Of Combinations

Quantity A: The number of potential combinations when three choices are made from seven options.

Quantity B: The number of potential combinations when four choices are made from seven options.

Possible Answers:

The two quantities are equal.

Quantity A is greater.

The relationship cannot be determined.

Quantity B is greater.

Correct answer:

The two quantities are equal.

Explanation:

With selections made from potential options, the total number of possible combinations (order doesn't matter) is:

$C(n,k)=\frac{n!}{(n-k)!k!}$

Quantity A:

$C(7,3)=\frac{7!}{(7-3)!3!}=\frac{7!}{(4)!3!}$

Quantity B:

$C(7,4)=\frac{7!}{(7-4)!4!}=\frac{7!}{(3)!4!}$

It is not necessary to perform this calculation, (although the value for each is thirty-five) and on the GRE needless calculations should be avoided.

The two quantities are equal.

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GRE Math : Arithmetic

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #2 : Combinations

Example Question #11 : How To Find The Greatest Or Least Number Of Combinations

Example Question #3 : Combinations

Example Question #3 : Combinations

Example Question #331 : Arithmetic

Example Question #4 : Combinations

Example Question #24 : How To Find The Greatest Or Least Number Of Combinations

Example Question #83 : Probability & Statistics

Example Question #331 : Arithmetic

Example Question #21 : How To Find The Greatest Or Least Number Of Combinations

All GRE Math Resources

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