## Common Scenarios

Permutation problems on the GRE can appear in a variety of forms. Below are two examples of common variations.

## Example

In how many arrangements can 5 people sit on a 5-person bench?

### Solution

In this scenario, there cannot be repeating values. If Bob is sitting in the first seat, he can’t also sit in the second seat. So this is a permutation without replacement.

**Method 1**

**1. Figure out how many places there are to fill.**

There are 5 seats on the bench: ___ ___ ___ ___ ___

**2. Figure out how many objects can potentially go into each place.**

The number of available people decreases as people are selected: __ 5 __ __ 4 __ __ 3 __ __ 2 __ __ 1 __

**3. Multiply.**

__5 __ × __ 4 __ × __ 3 __ × __ 2 __ × __ 1 __ = 5! = 120

**Method 2**

5 *P* 5 = \dfrac{5!}{(5 \,-\, 5)!} = \dfrac{5!}{0!} = 5! = 120

Notice that 0! = 1.

## Example

There are six daily dinner options in the cafeteria at John’s university. John is not allowed to order any meal a second time until he has ordered all the meals once. How many different ordering options does John have for the first four days?

### Solution

This may appear to be different from the bench example, but it is conceptually identical. John cannot be served the same meal twice before trying all the other meals, so he has four days of decreasing possibilities. It is another non-replacement permutation.

*Method 1*

**1. Figure out how many places there are to fill.**

John has four days of meals: ___ ___ ___ ___

**2. Figure out how many objects can potentially go into each place.**

On the first day there are 6 meals available, and the number decreases as meals are selected:

__ 6 __ __ 5 __ __ 4 __ __ 3 __

**3. Multiply.**

__6 __ × __ 5 __ × __ 4 __ × __ 3 __ = 30 × 12 = 360

*Method 2*

6 *P* 4 = \dfrac{6!}{(6 – 4)!} = \dfrac{6!}{2!}

= \dfrac{6 × 5 × 4 × 3 × 2!}{2!}

= 6 × 5 × 4 × 3 = 360

## Order with Restrictions

Permutation problems may have restrictions on what objects can go in each place, or on which objects can be next to each other.

## Example

In how many ways can a pet shop line up 3 cats and 3 dogs in 6 cages if the cats must be in the second, fourth, and sixth cages?

### Solution

This is a non-replacement permutation.

*Method 1*

**1. Figure out how many places there are to fill.**

There are six cages for the animals, so there are six places:

___ ___ ___ ___ ___ ___

**2. Figure out how many objects can potentially go into each place.**

This step is where things change slightly. Visualize the restrictions. There are 3 cats and 3 dogs, but the cats must be in the second, fourth, and sixth cages. That means the dogs must be in the first, third, and fifth cages: __ 3 __ __ 3 __ __ 2 __ __ 2 __ __ 1 __ __ 1 __

**3. Multiply.**

__3 __ × __ 3 __ × __ 2 __ × __ 2 __ × __ 1 __ × __ 1 __ = 36

*Method 2*

This question has 2 permutations: one for the dogs and one for the cats. Find each permutation then multiply.

There are 3 dogs and 3 cats, so each has the same formula:

3 P 3 = \dfrac{3!}{(3 \,-\, 3)!} = 3! = 6

So there are 6 order possibilities for the dogs and 6 for the cats.

The total number of permutations is 6 × 6 = 36.

**Example**

## Permutations with Repetitions

If you consider the letters D A D to be three different letters, then there are 6 permutations of their order.

**Method 1:** 3 × 2 × 1 = 6

**Method 2:** 3 *P* 3 = 6

**Method 3: **D A D, D A D, A D D,

A D D, D D A, D D A

If you consider the letters **D A D** to only have 2 different letters, then there are just 3 distinguishable permutations of their order.

**D A D A D D D D A**

The formula for permutations that have repeating objects is:

\dfrac{n!}{\textit{s}_{\displaystyle{1}}! × \textit{s}_{\displaystyle{2}}! × ...}\\[3ex]where ** n** is the number of objects and

**is the number of times each object is repeated.**

*s*

ExampleFind the number of distinguishable permutations for the letters.

a) DOODLE

b) DECEMBER

### Solution

**a)** There are 6 letters. The letters D and O repeat.

\dfrac{n!}{\textit{s}_{\displaystyle{1}}! × \textit{s}_{\displaystyle{2}}!} = \dfrac{6!}{2! × 2!}

= \dfrac{6 × 5 × 4 × 3 × 2 × 1}{2 × 1 × 2 × 1} = 180

**b)** There are 8 letters. The letter E repeats 3 times.

\dfrac{n!}{\textit{s}!} = \dfrac{8!}{3!} = \dfrac{8 × 7 × 6 × 5 × 4 × 3!}{3!}

= 8 × 7 × 6 × 5 × 4 = 56 × 120 = 6720

Notice that 8! has 3! as a factor.

Note: Permutations of letters are called anagrams if the new order of the letters is also a word. For example, DAD is an anagram of ADD; and SILENT is an anagram of LISTEN. But when counting permutations of letters, each permutation does not need to form a word.

*To see further discussion of the above example, click here.*

## Example

A gardener has 6 pots but only 4 different plants to be potted. How many arrangements are possible for all 6 pots?

### Solution

There are 6 pots. There are 4 pots that have a different plant (plant A, plant B, plant C, plant D). But the 2 pots without a plant are “repeats” (no plant, no plant).

\dfrac{n!}{\textit{s}!} = \dfrac{6!}{2!} = \dfrac{6 × 5 × 4 × 3 × 2 × 1}{2 × 1}

= 6 × 5 × 4 × 3 = 360

## 800score Tip:

Keep in mind that the GRE’s effectiveness relies on its ability to not be “beaten” by standard test preparation.So the GRE occasionally designs trick questions to penalize over-preparation. Questions about pulling cards or marbles may not always be the typical “cards” or “marble” questions you are used to seeing. In other words, they may not be permutation questions.

Always pay close attention to what the question is actually asking, and don’t make assumptions.

## Example

A company is creating different ID numbers for its employees. Senior-level employees will receive 4-digit ID numbers, and junior-level employees will receive 5-digit ID numbers. The first digit of any ID number cannot be zero, and no digits will be repeated in any ID number. What is the ratio of the total number of possible senior-level ID numbers to the total number of possible junior-level ID numbers?

### Solution

This question has 2 permutations: one for senior-level ID numbers and one for junior-level ID numbers. Find each permutation separately, and then make a ratio of the totals.

*For senior-level employees:*

**1. Figure out how many places there are to fill.**

There are 4 digits: ___ ___ ___ ___

**2. Figure out how many objects can potentially go into each place.**

The first digit can be any digit except 0. That means there are 9 options.

The second one can have any digit including 0, but it cannot have the same digit as the one before it, so there are also 9 options.

With two digits used, there are 8 options for the third digit, then 7 options for the fourth digit.

__ 9 __ __ 9 __ __ 8 __ __ 7 __

**3. Multiply.**

__9 __ × __ 9 __ × __ 8 __ × __ 7 __

*But don’t complete the calculation yet*. As always, keep in factored form and watch for cancellation.

*For junior-level employees:*

The logic is the same here, except there are 5 digits. The final step will be 9 × 9 × 8 × 7 × 6.

Finally, the question wants the ratio of these two permutations.

\dfrac{9 × 9 × 8 × 7}{9 × 9 × 8 × 7 × 6} = \dfrac{1}{6}\\[1ex]All the numbers cancel out except for 6. The result is simply 1/6.