GRE Math : Permutation / Combination

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Example Questions

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 #42 : Permutation / Combination

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 #41 : Permutation / Combination

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 #44 : Permutation / Combination

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:

Quantity B is greater.

Quantity A is greater.

The relationship cannot be determined.

The two quantities are equal.

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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Example Question #43 : Permutation / Combination

Bryant is purchasing basketballs from the basketball store. If he is purchasing three basketballs and there are seven basketballs to choose from, how many different combinations of basketballs can he buy?

Possible Answers:

210

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

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

Since Bryant is buying three basketballs from a selection of seven, the number of possible combinations is

$C(7,3)=\frac{7!}{(7-3)!3!}=35$

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Example Question #46 : Permutation / Combination

At a restaurant there is a choice of ten entrees and six potential sides. If an entree comes with two unique sides, how many different dinner options are possible?

Possible Answers:

560

300

3360

150

Correct answer:

150

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

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

One option is chosen from the ten entrees, so the number of entree combinations is

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

Two options are chosen from the six sides, so the number of side combinations is

$C(6,2)=\frac{6!}{(6-2)!2!}=15$

The total number of combinations is the product of these individual results:

10(15)

150

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GRE Math : Permutation / Combination

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #41 : Permutation / Combination

Example Question #42 : Permutation / Combination

Example Question #41 : Permutation / Combination

Example Question #42 : Permutation / Combination

Example Question #43 : Permutation / Combination

Example Question #44 : Permutation / Combination

Example Question #43 : Permutation / Combination

Example Question #46 : Permutation / Combination

All GRE Math Resources

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