Order is Important

A permutation is an arrangement of objects or elements in a specific order. Sometimes it is the ordering of all the elements in a group. Often it is the order of choosing elements from a group. For example, most padlocks use three numbers between 0 and 99, but it is the order of the numbers that let’s you open the lock.

There are 3 steps to find permutations:

1. Figure out how many places there are to fill.
2. Figure out how many objects can potentially go into each place.
3. Multiply.

Example

You roll a 6-sided die and pull a card from a deck of 52 distinct cards. How many different pairs of outcomes are possible?
(Note: die is the singular of dice.)

Solution

1. Figure out how many places there are to fill.
You are rolling a die and pulling a card, so there are 2 places: ___ ___

2. Figure out how many objects can potentially go into each place.
Rolling a die has 6 different outcomes and pulling a card has 52 outcomes: _6_ _52_

3. Multiply.
_6_ × _52_ = 312 There are 312 different possible outcomes.

Example

In Country X, three digit area codes are being given to each town. The first digit will be any number from 2 to 9, inclusive, the second digit can only be either 0 or 1, and the third digit can be any number from 0 to 9, inclusive. How many different area codes can be issued in Country X?

Solution

1. Figure out how many places there are to fill.
There are 3 digits, so there are 3 places to fill: ___ ___ ___

2. Figure out how many objects can potentially go into each place.
The first digit will be any number from 2 to 9, so there are 8 options.
The second digit can only be either 0 or 1, so there are 2 options.
The third digit can be any number from 0 to 9, so there are 10 options.
_8_ _2_ _10_

3. Multiply.
_8_ × _2_ × _10_ = 160 There are 160 possible area codes.

Permutations Without Replacement

It doesn’t matter how many times you roll a fair 6-sided die, there will be 6 possible outcomes for each roll. In many permutation questions, however, the number of possibilities will change with each placement or event. These are called permutations without replacement.

Example

A student wants to put 3 of her 7 different books into the empty spaces on the bookshelf. How many different possibilities are there for the order of the 3 books on the shelf?

Solution

1. Figure out how many places there are to fill.
She is choosing 3 books: ___ ___ ___

2. Figure out how many objects can potentially go into each place.
There are 7 possible books to choose from to fill the first space, 6 books for the second space and 5 books for the third space.
_7_ _6_ _5_

3. Multiply.
_7_ × _6_ × _5_ = 210 There are 210 possible orders for putting the 3 books on the shelf.

Permutations Formula

The formula for non-replacement permutations follows from the steps used above.

n P r = n!/(n – r)!
In this formula:
n stands for the number of distinct objects that you are choosing from.
r stands for the number of objects being chosen, or the places to fill.
P stands for permutation, and is not an arithmetic part of the equation.
The exclamation point (!) denotes the factorial of that number.
Remember that factorial n! means the product of all integers from n to 1.

3! = 3 × 2 × 1 = 6
5! = 5 × 4 × 3 × 2 × 1 = 120

Example

A student wants to put 3 of her 7 different books into the empty spaces on the bookshelf. How many different possibilities are there for the order of the 3 books on the shelf?

Solution

Solve this again, this time using the formula.

There are 7 possible books to choose from and 3 are being chosen.

7 P 3 = 7!/(7 – 3)! = 7!/4! = 7 × 6 × 5 × 4 × 3 × 2 × 1/4 × 3 × 2 × 1 = 7 × 6 × 5 = 210

Example

Sally is given a bag with 6 tiles, each numbered with a different digit from 1 to 6. She pulls out 3 tiles, one at a time, and lays them down in that order. How many different 3-digit numbers could she create?

Solution

The number tiles are not returned to the bag, so this is permutations without replacement.

Method 1

1. Figure out how many places there are to fill.
Pulling 3 tiles: ___ ___ ___

2. Figure out how many objects can potentially go into each place.
She starts with 6 tiles and that number decreases as tiles are selected: _6_ _5_ _4_

3. Multiply.
_6_ × _5_ × _4_ = 120

Method 2

6P 3 = 6!/(6 – 3)! = 6!/3! = 6 × 5 × 4 × 3 × 2 × 1 / 3 × 2 × 1 = 6 × 5 × 4 = 120

Notice that 6! has 3! as a factor.

800score Tip:

You can decide whether you prefer using logic or the formula. However, if you choose to use the formula it is important to also understand the logic of permutations in order to maximize your score. Remember, the GRE tests your ability to solve problems more than your ability to memorize formulas.

Comparing With or Without Replacement

On the GMAT you will need to be able to distinguish permutations with replacement from permutations without replacement. Below is a comparison to help you recognize each type.

With Replacement

Potential outcomes that are constant.

Multiply the number of possible outcomes.

Without Replacement

Potential outcomes decrease with each selection.

Use logic or the formula.

Rolling a 6-sided die

How many possibilities in 5 rolls?

6 × 6 × 6 × 6 × 6

Rolling a 6-sided die with no repeats

How many possibilities in 5 rolls with no repeated values?

6 × 5 × 4 × 3 × 2

Pulling a marble from a bag then putting it back

A marble is pulled from a bag and then put back in the bag. If the bag contains 9 green marbles and 9 blue marbles, how many different orders of 4 marbles are possible?

2 × 2 × 2 × 2

Pulling marbles from a bag

A bag contains 9 green marbles and 9 blue marbles. If 4 marbles are pulled, how many different orders of marbles are possible?

18P 4=18!=18!
(18 – 4)!14!

Password or Code

A code with 3 numbers is selected from 40 numbers. How many possible codes are there?

40 × 40 × 40

Password or Code where no number is repeated

A code with 3 numbers is selected from 40 numbers, and no number can be used twice. How many possible codes are there?

40 × 39 × 38

Pulling a card from a deck with 52 distinct cards and putting it back

A card is drawn from a deck then replaced and the deck is shuffled. How many different groups of cards (hands) can be formed with 4 draws?

52 × 52 × 52 × 52

Pulling from a deck of 52 distinct cards.

Four cards are drawn from a deck of cards. How many different groups of cards (hands) can be formed?

52 × 51 × 50 × 49

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