GRE Math : Permutation / Combination

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Example Questions

Example Question #21 : Permutation / Combination

The President has to choose from six members of congress to serve on a committee of three possible members. How many different groups of three could he choose?

Possible Answers:

120

Correct answer:

Explanation:

This is a combinations problem which means order does not matter. For his first choice the president can choose from 6, the second 5, and the third 4 so you may think the answer is 6 * 5 * 4, or 120; however this would be the answer if he were choosing an ordered set like vice president, secretary of state, and chief of staff. In this case order does not matter, so you must divide the 6 * 5 * 4 by 3 * 2 * 1, for the three seats he’s choosing. The answer is 120/6, or 20.

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Example Question #21 : Combinational Analysis

Mohammed is being treated to ice cream for his birthday, and he's allowed to build a three-scoop sundae from any of the thirty-one available flavors, with the only condition being that each of these flavors be unique. He's also allowed to pick different toppings of the available , although he's already decided well in advance that one of them is going to be peanut butter cup pieces.

Knowing these details, how many sundae combinations are available?

Possible Answers:

2427300

202275

40455

242730

44950

Correct answer:

40455

Explanation:

Because order is not important in this problem (i.e. chocolate chip, pecan, butterscotch is no different than pecan, butterscotch, chocolate chip), it is dealing with combinations rather than permutations.

The formula for a combination is given as:

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

where is the number of options and is the size of the combination.

For the ice cream choices, there are thirty-one options to build a three-scoop sundae. So, the number of ice cream combinations is given as:

$C(31,3)=\frac{31!}{(31-3)!3!}=4495$

Now, for the topping combinations, we are told there are ten options and that Mohammed is allowed to pick two items; however, we are also told that Mohammed has already chosen one, so this leaves nine options with one item being selected:

$C(9,1)=\frac{9!}{(9-1)!1!}=9$

So there are 9 "combinations" (using the word a bit loosely) available for the toppings. This is perhaps intuitive, but it's worth doing the math.

Now, to find the total sundae combinations—ice cream and toppings both—we multiply these two totals:

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Example Question #1 : Combinations

If there are students in a class and people are randomly choosen to become class representatives, how many different ways can the representatives be chosen?

Possible Answers:

200

210

190

180

222

Correct answer:

190

Explanation:

To solve this problem, we must understand the concept of combination/permutations. A combination is used when the order doesn't matter while a permutation is used when order matters. In this problem, the two class representatives are randomly chosen, therefore it doesn't matter what order the representative is chosen in, the end result is the same. The general formula for combinations is $C(n,k)=\frac{n!}{(n-k)!k!}$ , where is the number of things you have and is the things you want to combine.

Plugging in choosing 2 people from a group of 20, we find

$C(20,2)=\frac{20!}{(20-2)!2!}=190$ . Therefore there are a 190 different ways to choose the class representatives.

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Example Question #1 : 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:

Quantity A is greater.

The relationship cannot be determined.

Quantity B is greater.

The two quantities are equal

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 #22 : Combinational Analysis

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:

Quantity B is greater.

Quantity A is greater.

The two quantities are equal.

The relationship cannot be established.

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 #2 : 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:

$\textup{The two quantities are equal}$

$\textup{Quantity A is greater}$

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

$\textup{Quantity B is greater}$

$\textup{The relationship cannot be determined}$

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 #21 : Permutation / Combination

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:

Quantity A is larger.

The two quantities are equal.

Quantity B is larger.

The relationship cannot be determined.

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 #22 : Permutation / Combination

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:

The two quantities are equal.

Quantity A is greater.

The relationship cannot be established.

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:

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

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 #22 : Permutation / Combination

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:

Quantity B is greater.

The two quantities are equal.

The relationship cannot be determined.

Quantity A 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 #22 : Permutation / Combination

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:

304

600

1200

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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GRE Math : Permutation / Combination

Study concepts, example questions & explanations for GRE Math

All GRE Math Resources

Example Questions

Example Question #21 : Permutation / Combination

Example Question #21 : Combinational Analysis

Example Question #1 : Combinations

Example Question #1 : Combinations

Example Question #22 : Combinational Analysis

Example Question #2 : Combinations

Example Question #21 : Permutation / Combination

Example Question #22 : Permutation / Combination

Example Question #22 : Permutation / Combination

Example Question #22 : Permutation / Combination

All GRE Math Resources

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