GRE Math : Permutation / Combination

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Example Question #31 : Permutation / Combination

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:

2820

245700

500

20475

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 #32 : Permutation / Combination

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 #33 : Permutation / Combination

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:

Quantity A is greater.

The two quantities are equal.

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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Example Question #32 : Permutation / Combination

Quantity A: The number of possible permutations when seven choices are made from ten options.

Quantity B: The number of possible permutations when five choices are made from eleven options.

Possible Answers:

Quantity A is greater.

The two quantities are equal.

Quantity B is greater.

The relationship cannot be determined.

Correct answer:

Quantity A is greater.

Explanation:

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

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

Quantity A:

$P(10,7)=\frac{10!}{(10-7)!}=604800$

Quantity B:

$P(11,5)=\frac{11!}{(11-5)!}=55440$

Quantity A is greater.

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Example Question #33 : Permutation / Combination

Quantity A: The number of combinations if four choices are made from eight options.

Quantity B: The number of combinations if five choices are made from eight options.

Possible Answers:

The relationship cannot be determined.

The two quantities are equal.

Quantity B is greater.

Quantity A is greater.

Correct answer:

Quantity A is greater.

Explanation:

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

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

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 #33 : Permutation / Combination

Jack is putting together his winter ensemble to take with him on a weekend ski trip. He'll be bringing two scarves out of a selection of twelve, four pairs of socks from a group of ten, and three sweaters from a choice of six. How many clothing combinations are available to him?

Possible Answers:

296

6906900

79833600

5292

277200

Correct answer:

277200

Explanation:

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

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

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

We'll be dealing with the potential combinations for the scarves, socks, and sweaters; the total amount of combinations will be the product of these three.

Scarves:

$C(12,2)=\frac{12!}{(12-2)!2!}=66$

Socks:

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

Sweaters:

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

The total number of combiations is

(66)(210)(20)=277200

He's certainly not hurting for choices.

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Example Question #4 : Permutations

Jill is picking out outfits for a three-day weekend, one for Friday, one for Saturday, and one for Sunday.

Fortunately all of her clothes match together really well, so she can be creative with her options, though she's decided each outfit is going to be a combination of blouse, skirt, and shoes.

She'll be picking from ten blouses, twelve skirts, and eight pairs of shoes. How many ways could her weekend ensemble be lined up?

Possible Answers:

1478400

319334400

396

2376

739200

Correct answer:

319334400

Explanation:

For this problem, order matters! Wearing a particular blouse on Friday is not the same as wearing it on Sunday. So that means that this problem will be dealing with permutations.

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

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

What we'll do is calculate the number of permutations for her blouses, skirts, and shoes seperately (determining how the Friday/Saturday/Sunday blouses/skirts/shoes could be decided), and then multiply these values.

Blouses:

$P(10,3)=\frac{10!}{(10-3)!}=720$

Skirts:

$P(12,3)=\frac{12!}{(12-3)!}=1320$

Shoes:

$P(8,3)=\frac{8!}{(8-3)!}=336$

Thus the number of potential outfit assignments is

720(1320)(336)=319334400

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Example Question #34 : Permutation / Combination

Sammy is at an ice cream shoppe, aiming to build a sundae from two different flavors from a choice of thirty-one, and three separate toppings from a choice of ten. How many kinds of sundaes can he make?

Possible Answers:

669600

749398

55800

1650

585

Correct answer:

55800

Explanation:

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

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

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

Sammy is making two sub combinations; one of ice cream and one of toppings. The total amount of combinations will be the product of these two.

Ice cream:

$C(31,2)=\frac{31!}{(31-2)!2!}=465$

Toppings:

$C(10,3)=\frac{10!}{(10-3)!3!}=120$

The total number of potential sundaes is

(465)(120)=55800

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Example Question #35 : Permutation / Combination

Jessie is on a shopping spree, and decides he wants to buy sweaters. If he elects to buy thirteen sweaters out of selection of forty-five, how many potential sweater combinations could he purchase?

Possible Answers:

4831832086

952210078682

328910671

73006209045

454611182252363136000

Correct answer:

73006209045

Explanation:

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

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

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

$C(45,13)=\frac{45!}{(45-13)!13!}=73006209045$

Jeez, Jessie, go easy.

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Example Question #36 : Permutation / Combination

Rachel is buying ice cream for a sundae. If there are twelve ice cream choices, how many scoops will give the maximum possible number of unique sundaes?

Possible Answers:

Correct answer:

Explanation:

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

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

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

In terms of finding the maximum number of combinations, the value of should be

$k_{max}=\frac{n_{even}}{2};k_{max}=\frac{n_{odd}\pm 1}{2}$

Since there are twelve options, a selection of six scoops will give the maximum number of combinations.

$C(12,6)=\frac{12!}{(12-6)!6!}=924$

C(12,7)=792

C(12,8)=495

C(12,9)=220

C(12,10)=66

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GRE Math : Permutation / Combination

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #31 : Permutation / Combination

Example Question #32 : Permutation / Combination

Example Question #33 : Permutation / Combination

Example Question #32 : Permutation / Combination

Example Question #33 : Permutation / Combination

Example Question #33 : Permutation / Combination

Example Question #4 : Permutations

Example Question #34 : Permutation / Combination

Example Question #35 : Permutation / Combination

Example Question #36 : Permutation / Combination

All GRE Math Resources

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