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Counting Principles

Solve counting problems using the Addition Principle.

Counting Principles

Solve counting problems using the addition principle.
Solve counting problems using the multiplication principle.

In probability theory, an outcome is a possible result of an experiment or trial.

In probability theory, an event is a set of outcomes of an experiment.

Disjoint events cannot happen at the same time. In other words, they are mutually exclusive.

The addition principle applies when we are making only one selection.

Example

Try it.

ⓐ Seven red and five green marbles are placed in a bag. How many marbles are there to choose from?

ⓑ Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a negative or an even number from A?

Solution

ⓐ There are 7 ways of picking a red marble and 5 ways of picking a green marble and we cannot pick a red and a green at the same time. Therefore, there are 7+5 = 12 ways of picking a marble.

ⓑ There are 3 negative numbers in A and 3 even numbers in A and the even numbers are not negative. Therefore, there are 3+3 = 6 ways of choosing a negative or an even number from A.

Solve counting problems using the addition principle.

Try it.

Ten red and six green marbles are placed in a bag. How many marbles are there to choose from?

Try it.

Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a positive or an odd number from A?

Try it.

A young boy is deciding on a snack for the afternoon. He is deciding among 5 different chips, 3 different fruits and 2 different vegetables. What is the total number of snack options?

Condensed — the full section is in OpenStax College Algebra 2e.

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, as we can see in .

Example

Try it.

There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options?

Solution

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.

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 in .

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

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.

\[\begin{array}{llll}\#\ \text{of appetizer options }\times & \#\ \text{of entree options }\times & \#\ \text{of dessert options} & \\ \ 2\ \times & \ 3\ \times & \ 2 & =12\end{array}\]

Example

Try it.

Diane packed 2 skirts, 4 blouses, and 2 sweaters 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.

Solution

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.

Finding the Number of Permutations of

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.

Condensed — the full section is in OpenStax College Algebra 2e.

Find the Number of Combinations Using the Formula

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 \(r\) objects from a set of \(n\) objects where the order does not matter can be written as \(C(n,r).\) Just as with permutations, \(\text{C}(n,r)\) can also be written as \({\ }_{n}{C}_{r}.\) In this case, the general formula is as follows.

\[\text{C}(n,r)=\frac{n!}{r!(n-r)!}\]

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 \(3!=3\cdot 2\cdot 1=6\) ways to order 3 paintings. There are \(\frac{24}{6},\) 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.

Example

Try it.

A fast food restaurant offers five side dish options. Your meal comes with two side dishes.

ⓐHow many ways can you select your side dishes?
ⓑHow many ways can you select 3 side dishes?

Solution

ⓐWe want to choose 2 side dishes from 5 options. \[\text{C}(5,2)=\frac{5!}{2!(5-2)!}=10\]
ⓑWe want to choose 3 side dishes from 5 options. \[\text{C}(5,3)=\frac{5!}{3!(5-3)!}=10\]

Finding the Number of Subsets of a Set

We have looked only at combination problems in which we chose exactly \(r\) 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 \(C(5,0)=1\) way to order a pizza with no toppings. There are \(C(5,1)=5\) ways to order a pizza with exactly one topping. If we continue this process, we get

\[C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)=32\]

There are 32 possible pizzas. This result is equal to \({2}^{5}.\)

We are presented with a sequence of choices. For each of the \(n\) objects we have two choices: include it in the subset or not. So for the whole subset we have made \(n\) choices, each with two options. So there are a total of \(2\cdot 2\cdot 2\cdot \ldots \cdot 2\) 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.

Example

Try it.

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?

Solution

We are looking for the number of subsets of a set with 4 objects. Substitute \(n=4\) into the formula.

\[\begin{array}{l}{2}^{n}={2}^{4} \\ \ =16\end{array}\]

There are 16 possible ways to order a potato.

Finding the Number of Permutations of

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 \(12!\) 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 \(12!\) permutations we counted are duplicates. The general formula for this situation is as follows.

\[\frac{n!}{{r}_{1}!{r}_{2}!\ldots {r}_{k}!}\]

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 \(4!\) ways to order the stars and \(3!\) ways to order the moon.

\[\frac{12!}{4!3!}=3\text{,}326\text{,}400\]

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

Example

Try it.

Find the number of rearrangements of the letters in the word DISTINCT.

Solution

There are 8 letters. Both I and T are repeated 2 times. Substitute \(n=8,{r}_{1}=2,\) and \({r}_{2}=2\) into the formula.

\[\frac{8!}{2!2!}=10\text{,}080\]

There are 10,080 arrangements.

Key Equations

number of permutations of \(n\) distinct objects taken \(r\) at a time	\(P(n,r)=\frac{n!}{(n-r)!}\)
number of combinations of \(n\) distinct objects taken \(r\) at a time	\(C(n,r)=\frac{n!}{r!(n-r)!}\)
number of permutations of \(n\) non-distinct objects	\(\frac{n!}{{r}_{1}!{r}_{2}!\ldots {r}_{k}!}\)

Key Concepts

If one event can occur in \(m\) ways and a second event with no common outcomes can occur in \(n\) ways, then the first or second event can occur in \(m+n\) ways. See .
If one event can occur in \(m\) ways and a second event can occur in \(n\) ways after the first event has occurred, then the two events can occur in \(m\times n\) ways. See .
A permutation is an ordering of \(n\) objects.
If we have a set of \(n\) objects and we want to choose \(r\) objects from the set in order, we write \(P(n,r).\)
Permutation problems can be solved using the Multiplication Principle or the formula for \(P(n,r).\) See and .
A selection of objects where the order does not matter is a combination.
Given \(n\) distinct objects, the number of ways to select \(r\) objects from the set is \(\text{C}(n,r)\) and can be found using a formula. See .
A set containing \(n\) distinct objects has \({2}^{n}\) subsets. See .
For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See .

Counting Principles

Solve counting problems using the addition principle.
Solve counting problems using the multiplication principle.

In probability theory, an outcome is a possible result of an experiment or trial.

In probability theory, an event is a set of outcomes of an experiment.

Disjoint events cannot happen at the same time. In other words, they are mutually exclusive.

The addition principle applies when we are making only one selection.

Example

Try it.

ⓐ Seven red and five green marbles are placed in a bag. How many marbles are there to choose from?

ⓑ Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a negative or an even number from A?

Solution

ⓐ There are 7 ways of picking a red marble and 5 ways of picking a green marble and we cannot pick a red and a green at the same time. Therefore, there are 7+5 = 12 ways of picking a marble.

ⓑ There are 3 negative numbers in A and 3 even numbers in A and the even numbers are not negative. Therefore, there are 3+3 = 6 ways of choosing a negative or an even number from A.

Solve counting problems using the addition principle.

Try it.

Ten red and six green marbles are placed in a bag. How many marbles are there to choose from?

Try it.

Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a positive or an odd number from A?

Try it.

A young boy is deciding on a snack for the afternoon. He is deciding among 5 different chips, 3 different fruits and 2 different vegetables. What is the total number of snack options?

Condensed — the full section is in OpenStax Precalculus 2e.

Using the Addition Principle

Example

Try it.

There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options?

Solution

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.

Using the Multiplication Principle

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

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.

\[\begin{array}{llll}\#\ \text{of appetizer options }\times & \#\ \text{of entree options }\times & \#\ \text{of dessert options} & \\ \ 2\ \times & \ 3\ \times & \ 2 & =12\end{array}\]

Example

Try it.

Solution

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.

Finding the Number of Permutations of

Condensed — the full section is in OpenStax Precalculus 2e.

Find the Number of Combinations Using the Formula

\[\text{C}(n,r)=\frac{n!}{r!(n-r)!}\]

Example

Try it.

A fast food restaurant offers five side dish options. Your meal comes with two side dishes.

ⓐHow many ways can you select your side dishes?
ⓑHow many ways can you select 3 side dishes?

Solution

ⓐWe want to choose 2 side dishes from 5 options. \[\text{C}(5,2)=\frac{5!}{2!(5-2)!}=10\]
ⓑWe want to choose 3 side dishes from 5 options. \[\text{C}(5,3)=\frac{5!}{3!(5-3)!}=10\]

Finding the Number of Subsets of a Set

\[C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5)=32\]

There are 32 possible pizzas. This result is equal to \({2}^{5}.\)

Example

Try it.

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?

Solution

We are looking for the number of subsets of a set with 4 objects. Substitute \(n=4\) into the formula.

\[\begin{array}{l}{2}^{n}={2}^{4} \\ \ =16\end{array}\]

There are 16 possible ways to order a potato.

Finding the Number of Permutations of

\[\frac{n!}{{r}_{1}!{r}_{2}!\ldots {r}_{k}!}\]

\[\frac{12!}{4!3!}=3\text{,}326\text{,}400\]

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

Example

Try it.

Find the number of rearrangements of the letters in the word DISTINCT.

Solution

There are 8 letters. Both I and T are repeated 2 times. Substitute \(n=8,{r}_{1}=2,\) and \({r}_{2}=2\) into the formula.

\[\frac{8!}{2!2!}=10\text{,}080\]

There are 10,080 arrangements.

Key Equations

number of permutations of \(n\) distinct objects taken \(r\) at a time	\(P(n,r)=\frac{n!}{(n-r)!}\)
number of combinations of \(n\) distinct objects taken \(r\) at a time	\(C(n,r)=\frac{n!}{r!(n-r)!}\)
number of permutations of \(n\) non-distinct objects	\(\frac{n!}{{r}_{1}!{r}_{2}!\ldots {r}_{k}!}\)

Key Concepts

If one event can occur in \(m\) ways and a second event with no common outcomes can occur in \(n\) ways, then the first or second event can occur in \(m+n\) ways. See .
If one event can occur in \(m\) ways and a second event can occur in \(n\) ways after the first event has occurred, then the two events can occur in \(m\times n\) ways. See .
A permutation is an ordering of \(n\) objects.
If we have a set of \(n\) objects and we want to choose \(r\) objects from the set in order, we write \(P(n,r).\)
Permutation problems can be solved using the Multiplication Principle or the formula for \(P(n,r).\) See and .
A selection of objects where the order does not matter is a combination.
Given \(n\) distinct objects, the number of ways to select \(r\) objects from the set is \(\text{C}(n,r)\) and can be found using a formula. See .
A set containing \(n\) distinct objects has \({2}^{n}\) subsets. See .
For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

ⓐ Seven red and five green marbles are placed in a bag. How many marbles are there to choose from?
ⓑ Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a negative or an even number from A?
ⓒ A student is shopping for a new computer. He is deciding among 2 desktop computers and 3 laptop computers. What is the total number of computer options?

Revelar la respuesta

ⓐ There are 7 ways of picking a red marble and 5 ways of picking a green marble and we cannot pick a red and a green at the same time. Therefore, there are 7+5 = 12 ways of picking a marble.
ⓑ There are 3 negative numbers in A and 3 even numbers in A and the even numbers are not negative. Therefore, there are 3+3 = 6 ways of choosing a negative or an even number from A.
ⓒ There are 2 options for a desktop and 3 options for a laptop and the student is shopping for one computer. So, he cannot pick both. Therefore, there are 2+3 = 5 total computer options.
Ten red and six green marbles are placed in a bag. How many marbles are there to choose from?
Let the set A = {−5,−3,−1,2,3,4,5,6}. How many ways are there to choose a positive or an odd number from A?
A young boy is deciding on a snack for the afternoon. He is deciding among 5 different chips, 3 different fruits and 2 different vegetables. What is the total number of snack options?
Solve counting problems using the multiplication principle.
- ⓐ Diane packed 2 skirts, 3 blouses, and 2 sweaters 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.
- ⓑ A restaurant offers a lunch special that includes an entree, a main dish, and a beverage. There are 3 types of entrees, 4 main dish options, and 5 beverage choices. Find the total number of possible lunch specials.
- ⓒ Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?
- ⓓ How many license plates consisting of 2 letters followed by 4 digits are possible?
Revelar la respuesta
- ⓐ There are 2 outcomes for the skirts, 3 outcomes for the blouses and 2 outcomes for the sweaters. The total number of possible outfits is then:
  \(\begin{array}{llllll}\underset{⏟}{2} & \cdot & \underset{⏟}{3} & \cdot & \underset{⏟}{2} & =12 \\ \text{Skirts} & & \text{Blouses} & & \text{Sweaters} & \end{array}\)
- ⓑ There are 3 outcomes for the entrees, 4 outcomes for the main dish and 5 outcomes for the beverages. The total number of lunch specials is:
  \(\begin{array}{llllll}\underset{⏟}{3} & \cdot & \underset{⏟}{4} & \cdot & \underset{⏟}{5} & =60 \\ \text{Entrees} & & Main Dish & & \text{Beverages} & \end{array}\)
- ⓒ There are 4 outcomes for the science class, 3 outcomes for the math class, 2 outcomes for the history class, and 3 outcomes for the English class. The total number of 4-course selections is:
  \(\begin{array}{llllllll}\underset{⏟}{4} & \cdot & \underset{⏟}{3} & \cdot & \underset{⏟}{2} & \cdot & \underset{⏟}{3} & =72 \\ \text{Science} & & \text{Math} & & \text{English} & & \text{History} & \end{array}\)
- ⓓ There are 26 outcomes for the 1st letter, 26 outcomes for the second letter, 10 outcomes for the first digit, 10 outcomes for the second digit, 10 outcomes for the third digit, and 10 outcomes for the 4th digit. The number of license plates is:
  \(\begin{array}{llllllllllll}\underset{⏟}{26} & \cdot & \underset{⏟}{26} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & \cdot & \underset{⏟}{10} & =6,760,000 \\ 1st letter & & 2nd letter & & 1st digit & & 2nd digit & & 3rd digit & & 4th digit & \end{array}\)
How many two-letter strings—the first letter from set A and the second letter from set B can be formed from the sets A = {b, c, d} and B = {a, e, i, o, u}?
If you have three types of meat to make a sandwich (turkey, roast beef, and ham), and two types of bread (wheat and rye), how many different sandwiches with one kind of meat can be created?
Next semester you are going to take one science class, one math class, one history class and one english class. According to the schedule you have 4 different science classes, 3 different math classes, 2 different history classes, and 3 different English classes to choose from. Assuming no scheduling conflicts, how many different four-course selections can you make?
In Missouri, license plates have 3 letters and 3 numbers. How many license plates, consisting of 3 letters followed by 3 digits are possible?
There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options?

Revelar la respuesta

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.
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?

Revelar la respuesta

7
Diane packed 2 skirts, 4 blouses, and 2 sweaters 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.

Revelar la respuesta

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.
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.

Revelar la respuesta

There are 60 possible breakfast specials.
At a swimming competition, nine swimmers compete in a race.
1. ⓐHow many ways can they place first, second, and third?
2. ⓑ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.)
3. ⓒHow many ways can all nine swimmers line up for a photo?
Revelar la respuesta
1. ⓐ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.
2. ⓑ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.
3. ⓒ
  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.
How many ways can the family line up for the portrait?

Revelar la respuesta

120
How many ways can the photographer line up 3 family members?

Revelar la respuesta

60
How many ways can the family line up for the portrait if the parents are required to stand on each end?

Revelar la respuesta

12
A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions?

Revelar la respuesta

Substitute \(n=12\) and \(r=9\) into the permutation formula and simplify.
\[\begin{array}{l}\ P(n,r)=\frac{n!}{(n-r)!}\ \\ P(12,9)=\frac{12!}{(12-9)!}=\frac{12!}{3!}=79\text{,}833\text{,}600\end{array}\]
There are 79,833,600 possible permutations of exam questions!
How many ways can the 7 actors line up?

Revelar la respuesta

\(P(7,7)=5,040\)
How many ways can 5 of the 7 actors be chosen to line up?

Revelar la respuesta

\(P(7,5)=2,520\)
A fast food restaurant offers five side dish options. Your meal comes with two side dishes.
1. ⓐHow many ways can you select your side dishes?
2. ⓑHow many ways can you select 3 side dishes?
Revelar la respuesta
1. ⓐWe want to choose 2 side dishes from 5 options. \[\text{C}(5,2)=\frac{5!}{2!(5-2)!}=10\]
2. ⓑWe want to choose 3 side dishes from 5 options. \[\text{C}(5,3)=\frac{5!}{3!(5-3)!}=10\]
An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?

Revelar la respuesta

\(C(10,3)=120\)
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?

Revelar la respuesta

We are looking for the number of subsets of a set with 4 objects. Substitute \(n=4\) into the formula.
\[\begin{array}{l}{2}^{n}={2}^{4} \\ \ =16\end{array}\]
There are 16 possible ways to order a potato.
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?

Revelar la respuesta

64 sundaes
Find the number of rearrangements of the letters in the word DISTINCT.

Revelar la respuesta

There are 8 letters. Both I and T are repeated 2 times. Substitute \(n=8,{r}_{1}=2,\) and \({r}_{2}=2\) into the formula.
\[\frac{8!}{2!2!}=10\text{,}080\]
There are 10,080 arrangements.
Find the number of rearrangements of the letters in the word CARRIER.

Revelar la respuesta

840
Use the Addition Principle of counting to explain how many ways event \(A\ \text{or }B\) can occur.

Revelar la respuesta

There are \(m+n\) ways for either event \(A\) or event \(B\) to occur.
Use the Multiplication Principle of counting to explain how many ways event \(A\ \text{and }B\) can occur.
When given two separate events, how do we know whether to apply the Addition Principle or the Multiplication Principle when calculating possible outcomes? What conjunctions may help to determine which operations to use?

Revelar la respuesta

The addition principle is applied when determining the total possible of outcomes of either event occurring. The multiplication principle is applied when determining the total possible outcomes of both events occurring. The word “or” usually implies an addition problem. The word “and” usually implies a multiplication problem.
Describe how the permutation of \(n\) objects differs from the permutation of choosing \(r\) objects from a set of \(n\) objects. Include how each is calculated.
What is the term for the arrangement that selects \(r\) objects from a set of \(n\) objects when the order of the \(r\) objects is not important? What is the formula for calculating the number of possible outcomes for this type of arrangement?

Revelar la respuesta

A combination; \(C(n,r)=\frac{n!}{(n-r)!r!}\)
Let the set \(A=\{-5,-3,-1,2,3,4,5,6\}.\) How many ways are there to choose a negative or an even number from \(A?\)
Let the set \(B=\{-23,-16,-7,-2,20,36,48,72\}.\) How many ways are there to choose a positive or an odd number from \(A?\)

Revelar la respuesta

\(4+2=6\)
How many ways are there to pick a red ace or a club from a standard card playing deck?
How many ways are there to pick a paint color from 5 shades of green, 4 shades of blue, or 7 shades of yellow?

Revelar la respuesta

\(5+4+7=16\)
How many outcomes are possible from tossing a pair of coins?
How many outcomes are possible from tossing a coin and rolling a 6-sided die?

Revelar la respuesta

\(2\times 6=12\)
How many two-letter strings—the first letter from \(A\) and the second letter from \(B—\) can be formed from the sets \(A=\{b,c,d\}\) and \(B=\{a,e,i,o,u\}?\)
How many ways are there to construct a string of 3 digits if numbers can be repeated?

Revelar la respuesta

\({10}^{3}=1000\)
How many ways are there to construct a string of 3 digits if numbers cannot be repeated?

Symbols used here

factorial
n × (n−1) × … × 1; the number of orderings of n things. 0! = 1.

imaginary unit
i² = −1.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

\binom{n}{k}

binomial coefficient, "n choose k"
Number of k-element subsets of n things: n!/(k!(n−k)!).

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

Euler's number
2.71828…, the base whose exponential is its own derivative.

\theta

theta
The usual name for an angle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbf{v},\ \vec{v}

vector
A quantity with magnitude and direction; a column of numbers.

How to: Counting Principles

Solve counting problems using the Addition Principle.
Solve counting problems using the Multiplication Principle.
Solve counting problems using permutations involving n distinct objects.
Solve counting problems using combinations.
Find the number of subsets of a given set.
Solve counting problems using permutations involving n non-distinct objects.
Solve counting problems using the addition principle.
Solve counting problems using the multiplication principle.

Questions people ask

What is a function, really?

A rule that assigns exactly one output to each input. The vertical-line test on a graph is the same idea: no input may have two outputs.

Why do we need complex numbers?

Because x² + 1 = 0 has no real solution, and allowing one new number i with i² = −1 makes every polynomial equation solvable. They then turn out to describe rotation, waves and alternating current more naturally than real numbers do.

Prueba tu propio

Parts of this page are adapted from OpenStax College Algebra 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Más en Precalculus

Complex numbers Polynomial functions Rational functions Sequences and series The binomial theorem Conic sections Vectors Exponential and logarithmic functions Polynomial division and the remainder theorem Parametric equations and polar coordinates