## Identify

Half the challenge in combination and permutation questions is identifying whether it is a combination or permutation. Being able to quickly distinguish the two is critical to a high score.

Here are some clues to look for.

#### Does order matter?

Does it matter if the items’ order or positions are changed? If so, it is a permutations problem; if not, it is a combinations problem.

#### Which words are used?

The exact terms “permutations” or combinations” will rarely be used on the SAT.

Some words or phrases that indicate permutations include:
arrange in a row, winning, letters, security codes, choosing in order, and choosing with/without replacement.

Some words or phrases that indicate combinations include:
handshakes, food, choosing from multiple groups, and choosing more than one person at the same time (such as friends or a team).

## Comparisons

Here are seven examples of similar permutations and combinations. Read and think through each one to understand the differences.

#### Permutations

Groups of elements or objects that require order, rank or positioning.

#### Combinations

Groups of elements or objects without order, rank or positioning.

Committee with position titles

The president is different than the vice-president.

Generic committee

There is no rank. The committee is just one homogenous body.

Layers of colors used in painting

There is a difference between blue being painted onto red, and red being painted onto blue.

Different colors mixed for painting

If the colors are being mixed before use, then it doesn’t matter which is poured into the mixing container first.

Security code or PIN

1435 is different from 5341.

Sum of the numbers in a PIN

1 + 4 + 3 + 5 is the same as
5 + 3 + 4 + 1.

Selecting clothes for events

Wearing your red shirt on Wednesday and your black shirt on Thursday is different than the reverse.

Packing clothes for a trip

When you’re packing your red shirt and black shirt for a trip it doesn’t matter the order you put them in the suitcase.

Competition rank

In a race, election or game, there is an obvious difference between placing first and third.

Selecting prizes

If you won a game at the carnival and can choose two stuffed animals, it doesn’t matter which one you choose first.

Arranging people in a row

Steve, Maria and John in a row is different than Maria, Steve, and John.

Choosing people for a class

If Steve, Maria and John are chosen to move into a new class together, it doesn’t matter who was picked first or last.

Saying Hello

You say hello to a friend and he/she says hello back. These are two separate events that can occur in either order.

Handshakes

If you and a friend shake hands you are sharing in one single act at one single moment, so there can be no order.

##### Example

Determine whether order matters for the following questions.

a) Marco is redecorating his apartment and wants to put up 2 paintings side by side. How many arrangements are possible if he has 6 paintings to choose from?

### Solution

a) Monet on the left and Escher on the right is a different room design than the reverse. Therefore, order matters. This is a permutations question.

##### Example

Determine whether order matters for the following questions.

b) Emma is purchasing six roses from a bouquet of 15. How many different ways can this be done?

### Solution

b) The 6 roses Emma purchases will be one group without classification. There is no order. This is a combinations problem.

##### Example

Determine whether order matters for the following questions.

c) Alex is giving 8 of his 20 books to 8 different family members. How many ways can he do this?

### Solution

c) Giving War and Peace to his father and Curious George to his niece is different than the other way around. Order matters. This is a permutations question.

##### Example

You have a box with 7 pieces of different types of chocolate candy. You plan to have 2 pieces for dessert. How many pairs of chocolates can you choose?

### Solution

You are choosing 2 chocolates and order does not matter (you are going to eat both chocolates, regardless of which you select first). This is a combinations question.

##### 1. Start the problem as if it was a permutations problem.

a. Figure out how many places there are to fill.
You are choosing 2 pieces of chocolate:  ___  ___

b. Figure out how many objects can potentially go into each place.
You are choosing from 7 pieces of chocolate. Each time a chocolate is selected, the number of chocolates in the box decreases:    7   6

c. Multiply.
7 × 6 = 42

There are 42 different orders possible for the 2 pieces of chocolate. But this is not the answer for the combinations question.

##### 2. Divide the answer by the factorial of the number of places.

There are 2 outcomes, so divide by 2! = 2.

42 / 2 = 21      There are 21 possible pairs of chocolates.

## Doing Calculations

#### 800score Tip:

On combinations, don’t do the multiplication right away; keep results in factored form. The combinations formula usually creates fractions with factorials that can be cancelled, so that will be the first step.

##### Example

There are 15 toppings to choose from at a pizza restaurant. You plan to order a 4-topping pizza. How many different pizzas could you order?

A) 864
B) 1,365
C) 1,728
D) 2,730
E) 32,760

### Solution

The order of toppings on a pizza does not matter. (Does it matter if pepperoni goes on before mushrooms?) This is a combinations question.

Method 1

##### 1. Start the problem as if it was a permutations problem.

a. Figure out how many places there are to fill.
You are choosing 4 toppings, so there are 4 outcomes:  ___  ___  ___  ___

b. Figure out how many objects can potentially go into each place.
There are 15 toppings to choose from. The number of different toppings to choose from decreases after each choice: 15   14   13   12

c. Multiply.
15 × 14 × 13 × 12       Don’t calculate yet. Wait and cancel first.

##### 2. Divide the answer by the factorial of the number of places.

You are eliminating all the repeat pizzas and so you only count the distinct ones.

There are 4 outcomes, so divide by 4!

\dfrac{15 × 14 × 13 × 12}{4!}

= \dfrac{15 × 14 × 13 × 12}{4 × 3 × 2 × 1}

= \dfrac{15 × 2 × 7 × 13 × 12}{12 × 2} = 15 × 7 × 13

Remember that the SAT is not looking for long calculations. Look at the equation and the answers.
The last digit of 15 × 7 × 13 will be 5, since multiplying the last digits gives you the ones place of the answer.
5 × 7 × 3 = 105, and the only answer choice that ends in 5 is choice (B).

If you do the calculation, it will be
15 × 7 × 13 = 1365.

The correct answer is choice (B).

Method 2

Use the formula.

15 C 4 = \dfrac{15!}{4! (15 \,-\, 4)!} = \dfrac{15!}{4! × 11!}

= \dfrac{15 × 14 × 13 × 12 × 11!}{4 × 3 × 2 × 11!}

= \dfrac{15 × 2 × 7 × 13 × 12}{12 × 2} = 15 × 7 × 13

Notice that 15! has 11! as a factor.

The last digit of 15 × 7 × 13 will be 5, since multiplying the last digits gives you the ones place of the answer.
5 × 7 × 3 = 105, and the only answer choice that ends in 5 is choice (B).

The correct answer is choice (B).

##### Example

An abstract painter has 12 colors on her palette. If she decides to paint using exactly 5 colors, how many different combinations of colors can she have?

### Solution

This question uses the word “combinations” so it is clearly a combinations question.

Method 1

##### 1. Start the problem as if it was a permutations problem.

a. Figure out how many places there are to fill.
There are 5 colors so there are 5 spaces:  ___  ___  ___  ___  ___

b. Figure out how many objects can potentially go into each place.
There are 12 colors to choose from. The number of different colors to choose from decreases after each choice: 12   11   10   9   8

c. Multiply.
12 × 11 × 10 × 9 × 8       Don’t calculate yet. Wait and cancel first.

##### 2. Divide the answer by the factorial of the number of places.

There are 5 outcomes, so divide by 5!

\dfrac{12 × 11 × 10 × 9 × 8}{5!}

= \dfrac{12 × 11 × 10 × 9 × 8}{5 × 12 × 2}

= 11 × 9 × 8 = 792

There are 792 possible combinations.

Method 2

Use the formula.

12 C 5 = \dfrac{12!}{5! (12 \,-\, 5)!} = \dfrac{12!}{5! 7!}

= \dfrac{12 × 11 × 10 × 9 × 8 × 7!}{5 × 4 × 3 × 2 × 7!}

= \dfrac{12 × 11 × 10 × 9 × 8}{5 × 4 × 3 × 2}

= \dfrac{12 × 11 × 10 × 9 × 8}{5 × 12 × 2}

= 11 × 9 × 8 = 792

Notice that 12! has 7! as a factor.

There are 792 possible combinations.

##### Example

Five cards are drawn from a deck of 52 distinct cards. How many different 5-card groups are possible?

A) 250
B) 55,297
C) 2,598,960
D) 6,635,642
E) 311,875,200

### Solution

This question asks about choosing a group of cards without specifying order, so it is a combinations question.

Method 1

##### 1. Start the problem as if it was a permutations problem.

a. Figure out how many places there are to fill.
The hand is 5 cards so there are 5 spaces:  ___  ___  ___  ___  ___

b. Figure out how many objects can potentially go into each place.
There are 52 cards to choose from. The number of different cards to choose from decreases after each choice: 52   51   50   49   48

c. Multiply.
52 × 51 × 50 × 49 × 48       Don’t calculate yet. Wait and cancel first.

##### 2. Divide the answer by the factorial of the number of places.

There are 5 outcomes, so divide by 5!

\dfrac{52 × 51 × 50 × 49 × 48}{5!}

= \dfrac{52 × 51 × 50 × 49 × 48}{5 × 4 × 3 × 2 × 1}

The SAT would not expect you to do such a long calculation. When you look at the answers, you see that they vary widely, so you can just estimate the values.

\dfrac{50 × 50 × 50 × 50 × 50}{100}

= \dfrac{312,500,000}{100} = 3,125,000

The only answer close to this value is 2,598,960. The correct answer is choice (C).

Method 2

Use the formula.

52 C 5 = \dfrac{52!}{5! (52 \,-\, 5)!} = \dfrac{52!}{5! \,47!}

= \dfrac{52 × 51 × 50 × 49 × 48 × 47!}{5 × 4 × 3 × 2 × 47!}

= \dfrac{52 × 51 × 50 × 49 × 48}{120}

Estimate.

\dfrac{50 × 50 × 50 × 50 × 50}{100}

= \dfrac{312,500,000}{100} = 3,125,000

The only answer close to this value is 2,598,960. The correct answer is choice (C).