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.

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:

1. soup, chicken, cake
2. soup, chicken, pudding
3. soup, fish, cake
4. soup, fish, pudding
5. soup, steak, cake
6. soup, steak, pudding
7. salad, chicken, cake
8. salad, chicken, pudding
9. salad, fish, cake
10. salad, fish, pudding
11. salad, steak, cake
12. salad, steak, pudding

We can also find the total number of possible dinners by multiplying.

We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle.

The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation.

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.

Finding the Number of Permutations of n Distinct Objects Using a Formula

For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. 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]. Another way to write this is [latex]{}_{n}{P}_{r}[/latex], a notation commonly seen on computers and calculators. To calculate [latex]P\left(n,r\right)[/latex], we begin by finding [latex]n![/latex], the number of ways to line up all [latex]n[/latex] objects. We then divide by [latex]\left(n-r\right)![/latex] to cancel out the [latex]\left(n-r\right)[/latex] items that we do not wish to line up.

Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is [latex]6\times 5\times 4=120[/latex]. Using factorials, we get the same result.

[latex]\dfrac{6!}{3!}=\dfrac{6\cdot 5\cdot 4\cdot 3!}{3!}=6\cdot 5\cdot 4=120[/latex]

There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows.

[latex]P\left(n,r\right)=\dfrac{n!}{\left(n-r\right)!}[/latex]

Note that the formula stills works if we are choosing all [latex]n[/latex] objects and placing them in order. In that case we would be dividing by [latex]\left(n-n\right)![/latex] or [latex]0![/latex], which we said earlier is equal to 1. So the number of permutations of [latex]n[/latex] objects taken [latex]n[/latex] at a time is [latex]\frac{n!}{1}[/latex] or just [latex]n!\text{.}[/latex]

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]\dfrac{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]\dfrac{12!}{4!3!}=3\text{,}326\text{,}400[/latex]

There are 3,326,400 ways to order the sheet of stickers.

Combinations

So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When we are selecting objects and the order does not matter, we are dealing with combinations. A selection of [latex]r[/latex] objects from a set of [latex]n[/latex] objects where the order does not matter can be written as [latex]C\left(n,r\right)[/latex]. Just as with permutations, [latex]\text{C}\left(n,r\right)[/latex] can also be written as [latex]{}_{n}{C}_{r}[/latex]. In this case, the general formula is as follows.

[latex]\text{C}\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are [latex]3!=3\cdot 2\cdot 1=6[/latex] ways to order 3 paintings. There are [latex]\frac{24}{6}[/latex], or 4 ways to select 3 of the 4 paintings. This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings.

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.

Use the Binomial Theorem

Previously, we studied combinations. In the shortcut to finding [latex]{\left(x+y\right)}^{n}[/latex], we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation [latex]\left(\begin{gathered}n\\ r\end{gathered}\right)[/latex] instead of [latex]C\left(n,r\right)[/latex], but it can be calculated in the same way. So

[latex]\left(\begin{gathered}n\\ r\end{gathered}\right)=C\left(n,r\right)=\dfrac{n!}{r!\left(n-r\right)!}[/latex]

The combination [latex]\left(\begin{gathered}n\\ r\end{gathered}\right)[/latex] is called a binomial coefficient. An example of a binomial coefficient is [latex]\left(\begin{gathered}5\\ 2\end{gathered}\right)=C\left(5,2\right)=10[/latex].

When we expand [latex]{\left(x+y\right)}^{n}[/latex] by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand [latex]{\left(x+y\right)}^{52}[/latex], we might multiply [latex]\left(x+y\right)[/latex] by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

[latex]\begin{align}&{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2} \\ &{\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ &{\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \end{align}[/latex]

First, let’s examine the exponents. With each successive term, the exponent for [latex]x[/latex] decreases and the exponent for [latex]y[/latex] increases. The sum of the two exponents is [latex]n[/latex] for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

[latex]\left(\begin{gathered}n\\ 0\end{gathered}\right),\left(\begin{gathered}n\\ 1\end{gathered}\right),\left(\begin{gathered}n\\ 2\end{gathered}\right),\dots,\left(\begin{gathered}n\\ n\end{gathered}\right)[/latex].

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

[latex]\begin{align}{\left(x+y\right)}^{n}& =\sum\limits _{k=0}^{n}\left(\begin{gathered}n\\ k\end{gathered}\right){x}^{n-k}{y}^{k} \\ & ={x}^{n}+\left(\begin{gathered}n\\ 1\end{gathered}\right){x}^{n - 1}y+\left(\begin{gathered}{c}n\\ 2\end{gathered}\right){x}^{n - 2}{y}^{2}+\dots+\left(\begin{gathered}n\\ n - 1\end{gathered}\right)x{y}^{n - 1}+{y}^{n} \end{align}[/latex]

Another way to see the coefficients is to examine the expansion of a binomial in general form, [latex]x+y[/latex], to successive powers 1, 2, 3, and 4.

[latex]\begin{align}&{\left(x+y\right)}^{1}=x+y \\ &{\left(x+y\right)}^{2}={x}^{2}+2xy+{y}^{2} \\ &{\left(x+y\right)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ &{\left(x+y\right)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4} \end{align}[/latex]

Can you guess the next expansion for the binomial [latex]{\left(x+y\right)}^{5}?[/latex]

Above is an illustration the following:

• There are [latex]n+1[/latex] terms in the expansion of [latex]{\left(x+y\right)}^{n}[/latex].
• The degree (or sum of the exponents) for each term is [latex]n[/latex].
• The powers on [latex]x[/latex] begin with [latex]n[/latex] and decrease to 0.
• The powers on [latex]y[/latex] begin with 0 and increase to [latex]n[/latex].
• The coefficients are symmetric.

To determine the expansion on [latex]{\left(x+y\right)}^{5}[/latex], we see [latex]n=5[/latex], thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of [latex]x[/latex], the pattern is as follows:

• Introduce [latex]{x}^{5}[/latex], and then for each successive term reduce the exponent on [latex]x[/latex] by 1 until [latex]{x}^{0}=1[/latex] is reached.
• Introduce [latex]{y}^{0}=1[/latex], and then increase the exponent on [latex]y[/latex] by 1 until [latex]{y}^{5}[/latex] is reached.
  [latex]{x}^{5},{x}^{4}y,{x}^{3}{y}^{2},{x}^{2}{y}^{3},x{y}^{4},{y}^{5}[/latex]

The next expansion would be

[latex]{\left(x+y\right)}^{5}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}[/latex].

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal’s Triangle.

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3^rd row, flank the ends of the rows with 1’s, and add [latex]1+1[/latex] to find the middle number, 2. In the [latex]n\text{th}[/latex] row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

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{gathered}5\\ 1\end{gathered}\right){x}^{4}y+\left(\begin{gathered}5\\ 2\end{gathered}\right){x}^{3}{y}^{2}+\left(\begin{gathered}5\\ 3\end{gathered}\right){x}^{2}{y}^{3}+\left(\begin{gathered}{c}5\\ 4\end{gathered}\right)x{y}^{4}+{y}^{5}[/latex]

The second term is [latex]\left(\begin{gathered}5\\ 1\end{gathered}\right){x}^{4}y[/latex]. The third term is [latex]\left(\begin{gathered}5\\ 2\end{gathered}\right){x}^{3}{y}^{2}[/latex]. We can generalize this result.

[latex]\left(\begin{gathered}n\\ r\end{gathered}\right){x}^{n-r}{y}^{r}[/latex]

Key Equations

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{gathered}n\\ r\end{gathered}\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{gathered}n\\ r\end{gathered}\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 matters