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NY Regents Exam - Integrated Algebra: Help and Review24 chapters | 260 lessons

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Lesson Transcript

Instructor:
*David Liano*

After completing this lesson, you will be able to state the fundamental counting principle and use it to calculate the number of possible outcomes for multiple events. You will also be able to use the fundamental counting principle to determine the number of permutations of a distinct set of objects.

The **calculating of possible outcomes** is a process for determining the number of possible results for an event. There are various methods for conducting this process. The best way to calculate the number of possible outcomes of an event is dependent on the type of event and the structure of the event.

A simple example can be found in sports. In any baseball game, there are two possible outcomes: Team *A* wins or Team *B* wins. However, a possibility problem can be much more complex than determining the number of possible winners of an athletic competition between two teams.

In this lesson, we will discuss how the fundamental counting principle is used to count the number of possible outcomes for multiple events and to count the number of permutations for a distinct group of objects.

Let's say that we want to order an ice cream cone from a local ice cream shop. The shop offers four flavors of ice cream (vanilla, chocolate, pistachio, and mint) and two types of cones (regular and cinnamon). We need to determine how many different types of ice cream cones we can order. One way to solve our problem is to create a tree diagram.

There are four possibilities from just choosing an ice cream flavor. Once we choose an ice cream flavor, we need to choose a cone. There are two possible cones to choose from, so there are two possible combinations with each flavor. Therefore, the possible number of ice cream cones is 4 x 2 = 8. We can also see this in the tree diagram.

We just used the **fundamental counting principle**. This principle states that if there are *p* possibilities for one event and *q* possibilities for a second event, then the number of possibilities for both events is *p* x *q*. We can add additional events to this formula. Let's say that we can also choose a topping for our ice cream cone from among three choices of toppings. Then the possible number of ice cream cones is 4 x 2 x 3 = 24.

A famous example of the fundamental counting principle is the possible combination of letters and numbers for a vehicle license plate. Of course, each state has its own policy for the number and types of characters that can be put on a license plate issued in that respective state. Let's say that our plate has four numbers and two letters similar to the New York plate shown. The numbers come first and then the letters.

The numbers can range from 0 to 9 and the letters can be any letter of the alphabet. There are 10 possible outcomes for each of the first four characters and 26 possible outcomes for each of the last two characters. Using the fundamental counting principle, the possible combination of numbers and letters on our license plate is 10 x 10 x 10 x 10 x 26 x 26 = 6,760,000. Just imagine using a tree diagram to solve this problem.

A **permutation** is an arrangement, or ordering, of a set of objects. Let's say that we have a horse race of six horses and that we need to arrange these horses in the starting gate. We want to determine how many arrangements of the six horses are possible.

There are six possible horses that we can choose to put in the first stall. After selecting one of the horses for the first stall, five horses remain for selection into the second stall. Therefore, each horse that could be in the first stall can be followed by any of the five remaining horses. According to the fundamental counting principle, there are 6 x 5 = 30 permutations of horses for the first two stalls. But we still need to arrange the rest of the horses. For the third stall, there are 4 horses remaining to choose from and so on. This pattern continues until all the horses are placed in the stalls of the starting gate. The total number of permutations of the six horses is 6 x 5 x 4 x 3 x 2 x 1 = 720.

The starting gate problem is very similar to our earlier examples except that we keep selecting from the same set of objects, of which there is one less to choose from after each selection.

If we need to arrange only a portion of a set of objects, then we arrange the number of objects that have been directed. Let's say that we have seven distinct plants, but only need to arrange three of these plants on a shelf. We have seven possibilities for the first spot on the shelf. For the second spot on the shelf, we only have six possibilities because one of the plants has already been chosen. Finally, we have five possibilities for the third spot. Our final answer is 7 x 6 x 5 = 210. There are 210 possible ways to arrange 3 plants selected from a set of 7 distinct plants.

The **fundamental counting principle** is the primary rule for calculating the number of possible outcomes. If there are *p* possibilities for one event and *q* possibilities for a second event, then the number of possibilities for both events is *p* x *q*. If a third event is added such as *r*, then the number of possibilities for all three events is *p* x *q* x *r*. This formula can be extended to as many events as necessary. The fundamental counting principle is also helpful when determining the number of possible permutations of a set of objects. A **permutation** is an arrangement, or ordering, of a set of objects. We can think of each event as the remaining number of objects that are available to be arranged.

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NY Regents Exam - Integrated Algebra: Help and Review24 chapters | 260 lessons

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