## 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 lets you open the lock.

**There are 3 steps to find permutations:**

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

ExampleYou 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:dieis the singular ofdice.)

### 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.

ExampleIn 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 CountryX?

### 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.

ExampleA 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* = \dfrac{\textit{n}!}{(\textit{n} \,-\, \textit{r})!}

In this formula:

stands for the number of distinct objects that you are choosing from.*n*stands for the number of objects being chosen, or the places to fill.*r*stands for permutation, and is not an arithmetic part of the equation.*P*- 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

ExampleA 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* = \dfrac{7!}{(7 – 3)!} = \dfrac{7!}{4!}

= \dfrac{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*

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

= \dfrac{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 GMAT 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
Multiply the number of possible outcomes. |
## Without Replacement
Use logic or the formula. |

Rolling a 6-sided die
6 × 6 × 6 × 6 × 6 |
Rolling a 6-sided die with no repeats
6 × 5 × 4 × 3 × 2 |

Pulling a marble from a bag then putting it back
2 × 2 × 2 × 2 |
Pulling marbles from a bag
18 |

Password or Code
40 × 40 × 40 |
Password or Code where no number is repeated
40 × 39 × 38 |

Pulling a card from a deck with 52 distinct cards and putting it back
52 × 52 × 52 × 52 |
Pulling from a deck of 52 distinct cards.
52 × 51 × 50 × 49 |