GRE Math : How to find the greatest or least number of combinations

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Example Question #31 : How To Find The Greatest Or Least Number Of Combinations

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:

2376

739200

1478400

396

319334400

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 #32 : 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

585

1650

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 #33 : 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:

952210078682

4831832086

328910671

454611182252363136000

73006209045

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 #72 : Other Topics

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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Example Question #41 : Permutation / Combination

Claus is taking his twin brother Lucas out for ice cream. Claus knows that his brother is indecisive and wants to spend as little time choosing ice cream as possible. Claus can choose how many scoops Lucas can make for a sundae, as long as Lucas gets at least four. If there are twelve ice cream options, how many scoops should Claus tell Lucas to get?

Each scoop of ice cream is a unique flavor.

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

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

Once the number of choices goes above or below this value (or below the minimum kmax/above the maximum kmax for an odd number of max choices), the number of potential combinations decreases. The farther the value of from the max, the lower the amount of choices.

For this problem:

$k_{max}=\frac{12}{2}=6$

For the choices provided the greater difference from the max occurs for k=9 .

|9-6|=3

|8-6|=2

|7-6|=1

|6-6|=0

|5-6|=1

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Example Question #31 : How To Find The Greatest Or Least Number Of Combinations

Saito is purchasing model cars. If there are twenty-three cars available, a purchase of how many cars would offer the maximum number of combinations?

Possible Answers:

11.5

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 is an odd number of cars:

$k_{max}=\frac{23\pm 1}{2}=\frac{24}{2},\frac{22}{2}=11,12$

Of course, it is not possible to purchase half a set.

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Example Question #32 : How To Find The Greatest Or Least Number Of Combinations

Skater Blake is buying skateboards. If there are eleven skateboards to choose from, a purchase of how many skateboards would give the smallest number of potential combinations of the options given below?

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

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

Once the number of choices goes above or below this value (or below the smaller k_max/above the greater k_max for an odd number of total options), the number of potential combinations decreases. The farther the value of from the max, the lower the amount of choices.

In other words:

$|k-k{max}|_{up}\rightarrow combinations_{down}$

We're given an odd number of options so,

$k_{max}=\frac{11 \pm 1}{2}=5,6$

For the available choices 3, 4, 5, 6, or 7:

|3-5|=2

|4-5|=1

|5-5|=0

|6-6|=0

|7-6|=0

will give the minimum number of choices.

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Example Question #42 : Permutation / Combination

Marty is purchasing sweaters. If there are unique sweaters for sale, a purchase of how many sweaters from the options below will give the minimum amount of potential sweater combinations?

Possible Answers:

Correct answer:

Explanation:

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

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

The number of combinations increases the closer the value of is to $\frac{1}{2}n$ .

In the case of being even:

$k_{max}=\frac{n_{even}}{2}$

In the case of being odd:

$k_{max_{1,2}}=\frac{n_{odd} \pm 1}{2}$

When a value of drifts farther from these values, the number of potential combinations decreases to a minimum of .

$|k-k_{max}|_{up} \rightarrow #Combinations_{down}$

Note that for an odd , consider the difference small values of $(<k_{max_{lower}})$ and the smaller $k_{max}$ , and the difference of large values of $(>k_{max_{greater}})$ and the larger $k_{max}$ .

Since there are 37 options, an odd number:

$k_{max_{1,2}}=\frac{37 \pm 1}{2}=18,19$

For the potential numbers of purchased sweaters:

|16-18|=2

|17-18|=1

|18-18|=0

Note that nineteen also corresponds to the maximum number of possible combinations.

|19-19|=0

|20-19|=1

k=16 gives the smallest amount of potential combinations for the choices presented.

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Example Question #43 : Permutation / Combination

Clark is in the market for new capes. If the cape store sells 48 unique types of capes, a purchase of how many capes will correspond to the minimum amount of potential combinations of capes?

Possible Answers:

Correct answer:

Explanation:

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

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

The number of combinations increases the closer the value of is to $\frac{1}{2}n$ .

In the case of being even:

$k_{max}=\frac{n_{even}}{2}$

In the case of being odd:

$k_{max_{1,2}}=\frac{n_{odd} \pm 1}{2}$

When a value of drifts farther from these values, the number of potential combinations decreases to a minimum of .

$|k-k_{max}|_{up} \rightarrow #Combinations_{down}$

Note that for an odd , consider the difference small values of and the smaller $k_{max}$ , and the difference of large values of and the larger $k_{max}$ .

Since is even:

$k_{max}=\frac{48}{2}=24$

k:24;|24-24|=0;C(48,24)=32247603683100

k:25;|25-24|=1;C(48,25)=30957699535776

k:26;|26-24|=2;C(48,26)=27385657281648

k:27;|27-24|=3;C(48,27)=22314239266528

k:28;|28-24|=4;C(48,28)=16735679449896

is farthest from $k_{max}$ and gives the least amount of possible combinations.

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Example Question #931 : Gre Quantitative Reasoning

A sundae can be ordered with any of ten possible toppings.

Quantity A: The number of different possible sundaes with three toppings.

Quantity B: The number of different possible sundaes with seven toppings.

Possible Answers:

The relationship cannot be determined.

Quantity B is greater.

The two quantities are equal.

Quantity A is greater.

Correct answer:

The two quantities are equal.

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

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

Quantity A:

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

Quantity B:

$C(10,7)=\frac{10!}{(10-7)!7!}=120$

The two quantities are equal.

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GRE Math : How to find the greatest or least number of combinations

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #31 : How To Find The Greatest Or Least Number Of Combinations

Example Question #32 : Permutation / Combination

Example Question #33 : Permutation / Combination

Example Question #72 : Other Topics

Example Question #41 : Permutation / Combination

Example Question #31 : How To Find The Greatest Or Least Number Of Combinations

Example Question #32 : How To Find The Greatest Or Least Number Of Combinations

Example Question #42 : Permutation / Combination

Example Question #43 : Permutation / Combination

Example Question #931 : Gre Quantitative Reasoning

All GRE Math Resources

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