Counting Principles
A new company sells customizable cases for tablets and smartphones. Each case comes in a variety of colors and can be personalized for an additional fee with images or a monogram. A customer can choose not to personalize or could choose to have one, two, or three images or a monogram. The customer can choose the order of the images and the letters in the monogram. The company is working with an agency to develop a marketing campaign with a focus on the huge number of options they offer. Counting the possibilities is challenging! We encounter a wide variety of counting problems every day. There is a branch of mathematics devoted to the study of counting problems such as this one. Other applications of counting include secure passwords, horse racing outcomes, and college scheduling choices. We will examine this type of mathematics in this section.Permutations
Using the Addition Principle
The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options.
A General Note: The Addition Principle
According to the Addition Principle, if one event can occur in [latex]m[/latex] ways and a second event with no common outcomes can occur in [latex]n[/latex] ways, then the first or second event can occur in [latex]m+n[/latex] ways.Example: Using the Addition Principle
There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options?Answer:
We can add the number of vegetarian options to the number of meat options to find the total number of entrée options.
There are 7 total options.
Try It
A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop computers. What is the total number of computer options?Answer: 7
Using the Multiplication Principle
The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12 possible choices of one each as shown in the tree diagram.
The possible choices are:
- soup, chicken, cake
- soup, chicken, pudding
- soup, fish, cake
- soup, fish, pudding
- soup, steak, cake
- soup, steak, pudding
- salad, chicken, cake
- salad, chicken, pudding
- salad, fish, cake
- salad, fish, pudding
- salad, steak, cake
- salad, steak, pudding
A General Note: The Multiplication Principle
According to the Multiplication Principle, if one event can occur in [latex]m[/latex] ways and a second event can occur in [latex]n[/latex] ways after the first event has occurred, then the two events can occur in [latex]m\times n[/latex] ways. This is also known as the Fundamental Counting Principle.Example: Using the Multiplication Principle
Diane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits.Answer:
To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options.
There are 16 possible outfits.
Try It
A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials.Answer: There are 60 possible breakfast specials.
Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle
To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall.
There are four options for the first place, so we write a 4 on the first line.
After the first place has been filled, there are three options for the second place so we write a 3 on the second line.
After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product.
There are 24 possible permutations of the paintings.
How To: Given [latex]n[/latex] distinct options, determine how many permutations there are.
- Determine how many options there are for the first situation.
- Determine how many options are left for the second situation.
- Continue until all of the spots are filled.
- Multiply the numbers together.
Example: Finding the Number of Permutations Using the Multiplication Principle
At a swimming competition, nine swimmers compete in a race.- How many ways can they place first, second, and third?
- How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.)
- How many ways can all nine swimmers line up for a photo?
Answer:
- Draw lines for each place.
There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place.
Multiply to find that there are 504 ways for the swimmers to place. - Draw lines for describing each place.
We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place.
Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first. - Draw lines for describing each place in the photo.
There are 9 choices for the first spot, then 8 for the second, 7 for the third, 6 for the fourth, and so on until only 1 person remains for the last spot.
There are 362,880 possible permutations for the swimmers to line up.
Analysis of the Solution
Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of [latex]n[/latex] distinct objects can always be found by [latex]n![/latex].Try It
A family of five is having portraits taken. Use the Multiplication Principle to find the following. How many ways can the family line up for the portrait?Answer: 120
How many ways can the photographer line up 3 family members?Answer: 60
How many ways can the family line up for the portrait if the parents are required to stand on each end?Answer: 12
Finding the Number of Permutations of n Non-Distinct Objects
We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be [latex]12![/latex] ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the [latex]12![/latex] permutations we counted are duplicates. The general formula for this situation is as follows.[latex]\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}[/latex]
In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are [latex]4![/latex] ways to order the stars and [latex]3![/latex] ways to order the moon.[latex]\frac{12!}{4!3!}=3\text{,}326\text{,}400[/latex]
There are 3,326,400 ways to order the sheet of stickers.A General Note: Formula for Finding the Number of Permutations of n Non-Distinct Objects
If there are [latex]n[/latex] elements in a set and [latex]{r}_{1}[/latex] are alike, [latex]{r}_{2}[/latex] are alike, [latex]{r}_{3}[/latex] are alike, and so on through [latex]{r}_{k}[/latex], the number of permutations can be found by[latex]\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}[/latex]
Example: Finding the Number of Permutations of n Non-Distinct Objects
Find the number of rearrangements of the letters in the word DISTINCT.Answer: There are 8 letters. Both I and T are repeated 2 times. Substitute [latex]n=8, {r}_{1}=2, [/latex] and [latex] {r}_{2}=2 [/latex] into the formula.
[latex]\frac{8!}{2!2!}=10\text{,}080 [/latex]
There are 10,080 arrangements.Try It
Find the number of rearrangements of the letters in the word CARRIER.Answer: 840
Finding the Number of Subsets of a Set
We have looked only at combination problems in which we chose exactly [latex]r[/latex] objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible? To answer this question, we need to consider pizzas with any number of toppings. There is [latex]C\left(5,0\right)=1[/latex] way to order a pizza with no toppings. There are [latex]C\left(5,1\right)=5[/latex] ways to order a pizza with exactly one topping. If we continue this process, we get[latex]C\left(5,0\right)+C\left(5,1\right)+C\left(5,2\right)+C\left(5,3\right)+C\left(5,4\right)+C\left(5,5\right)=32[/latex]
There are 32 possible pizzas. This result is equal to [latex]{2}^{5}[/latex]. We are presented with a sequence of choices. For each of the [latex]n[/latex] objects we have two choices: include it in the subset or not. So for the whole subset we have made [latex]n[/latex] choices, each with two options. So there are a total of [latex]2\cdot 2\cdot 2\cdot \dots \cdot 2[/latex] possible resulting subsets, all the way from the empty subset, which we obtain when we say "no" each time, to the original set itself, which we obtain when we say "yes" each time.A General Note: Formula for the Number of Subsets of a Set
A set containing n distinct objects has [latex]{2}^{n}[/latex] subsets.Example: Finding the Number of Subsets of a Set
A restaurant offers butter, cheese, chives, and sour cream as toppings for a baked potato. How many different ways are there to order a potato?Answer: We are looking for the number of subsets of a set with 4 objects. Substitute [latex]n=4[/latex] into the formula.
[latex]\begin{array}{l}{2}^{n}={2}^{4}\hfill \\ \text{ }=16\hfill \end{array}[/latex]
There are 16 possible ways to order a potato.Try It
A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How many different sundaes are possible?Answer: 64 sundaes
Using the Binomial Theorem to Find a Single Term
Expanding a binomial with a high exponent such as [latex]{\left(x+2y\right)}^{16}[/latex] can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of [latex]{\left(x+y\right)}^{5}[/latex].[latex]{\left(x+y\right)}^{5}={x}^{5}+\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y+\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}+\left(\begin{array}{c}5\\ 3\end{array}\right){x}^{2}{y}^{3}+\left(\begin{array}{c}5\\ 4\end{array}\right)x{y}^{4}+{y}^{5}[/latex]
The second term is [latex]\left(\begin{array}{c}5\\ 1\end{array}\right){x}^{4}y[/latex]. The third term is [latex]\left(\begin{array}{c}5\\ 2\end{array}\right){x}^{3}{y}^{2}[/latex]. We can generalize this result.[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]
A General Note: The (r+1)th Term of a Binomial Expansion
The [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion of [latex]{\left(x+y\right)}^{n}[/latex] is:[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]
How To: Given a binomial, write a specific term without fully expanding.
- Determine the value of [latex]n[/latex] according to the exponent.
- Determine [latex]\left(r+1\right)[/latex].
- Determine [latex]r[/latex].
- Replace [latex]r[/latex] in the formula for the [latex]\left(r+1\right)\text{th}[/latex] term of the binomial expansion.
Example: Writing a Given Term of a Binomial Expansion
Find the tenth term of [latex]{\left(x+2y\right)}^{16}[/latex] without fully expanding the binomial.Answer: Because we are looking for the tenth term, [latex]r+1=10[/latex], we will use [latex]r=9[/latex] in our calculations.
[latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex]
[latex]\left(\begin{array}{c}16\\ 9\end{array}\right){x}^{16 - 9}{\left(2y\right)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}[/latex]
Try It
Find the sixth term of [latex]{\left(3x-y\right)}^{9}[/latex] without fully expanding the binomial.Answer: [latex]-10,206{x}^{4}{y}^{5}[/latex]
Key Equations
| number of permutations of [latex]n[/latex] distinct objects taken [latex]r[/latex] at a time | [latex]P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}[/latex] |
| number of combinations of [latex]n[/latex] distinct objects taken [latex]r[/latex] at a time | [latex]C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}[/latex] |
| number of permutations of [latex]n[/latex] non-distinct objects | [latex]\frac{n!}{{r}_{1}!{r}_{2}!\dots {r}_{k}!}[/latex] |
| Binomial Theorem | [latex]{\left(x+y\right)}^{n}=\sum _{k - 0}^{n}\left(\begin{array}{c}n\\ k\end{array}\right){x}^{n-k}{y}^{k}[/latex] |
| [latex]\left(r+1\right)th[/latex] term of a binomial expansion | [latex]\left(\begin{array}{c}n\\ r\end{array}\right){x}^{n-r}{y}^{r}[/latex] |
Key Concepts
- If one event can occur in [latex]m[/latex] ways and a second event with no common outcomes can occur in [latex]n[/latex] ways, then the first or second event can occur in [latex]m+n[/latex] ways.
- If one event can occur in [latex]m[/latex] ways and a second event can occur in [latex]n[/latex] ways after the first event has occurred, then the two events can occur in [latex]m\times n[/latex] ways.
- A permutation is an ordering of [latex]n[/latex] objects.
- If we have a set of [latex]n[/latex] objects and we want to choose [latex]r[/latex] objects from the set in order, we write [latex]P\left(n,r\right)[/latex].
- Permutation problems can be solved using the Multiplication Principle or the formula for [latex]P\left(n,r\right)[/latex].
- A selection of objects where the order does not matter is a combination.
- Given [latex]n[/latex] distinct objects, the number of ways to select [latex]r[/latex] objects from the set is [latex]\text{C}\left(n,r\right)[/latex] and can be found using a formula.
- A set containing [latex]n[/latex] distinct objects has [latex]{2}^{n}[/latex] subsets.
- For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations.
- [latex]\left(\begin{array}{c}n\\ r\end{array}\right)[/latex] is called a binomial coefficient and is equal to [latex]C\left(n,r\right)[/latex].
- The Binomial Theorem allows us to expand binomials without multiplying.
- We can find a given term of a binomial expansion without fully expanding the binomial.
Glossary
Addition Principle if one event can occur in [latex]m[/latex] ways and a second event with no common outcomes can occur in [latex]n[/latex] ways, then the first or second event can occur in [latex]m+n[/latex] ways binomial coefficient the number of ways to choose r objects from n objects where order does not matter; equivalent to [latex]C\left(n,r\right)[/latex], denoted [latex]\left(\begin{array}{c}n\\ r\end{array}\right)[/latex] binomial expansion the result of expanding [latex]{\left(x+y\right)}^{n}[/latex] by multiplying Binomial Theorem a formula that can be used to expand any binomial combination a selection of objects in which order does not matter Fundamental Counting Principle if one event can occur in [latex]m[/latex] ways and a second event can occur in [latex]n[/latex] ways after the first event has occurred, then the two events can occur in [latex]m\times n[/latex] ways; also known as the Multiplication Principle Multiplication Principle if one event can occur in [latex]m[/latex] ways and a second event can occur in [latex]n[/latex] ways after the first event has occurred, then the two events can occur in [latex]m\times n[/latex] ways; also known as the Fundamental Counting Principle permutation a selection of objects in which order mattersLicenses & Attributions
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