Lesson 13.5: Counting Principles

New Concept

Learn the fundamental rules for counting possibilities. We'll explore when to add choices (Addition Principle) versus when to multiply them (Multiplication Principle), and apply these skills to solve problems involving arrangements (permutations) and selections (combinations).

What’s next

Now that you have the foundation, you'll tackle interactive examples and practice cards to master applying these principles to solve complex counting problems.

Property

According to the Addition Principle, 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.

Examples

• A coffee shop offers 8 types of coffee and 5 types of tea. A customer can choose a drink in $8 + 5 = 13$ ways.
• A bookstore has 10 new fiction books and 6 new non-fiction books. You can choose one new book to read in $10 + 6 = 16$ ways.
• In a car lot, there are 12 sedans and 9 SUVs available. The total number of vehicle options is $12 + 9 = 21$.

Explanation

Think of this as the 'or' rule. When you have to choose from one group OR another, and the groups are separate (no overlap), you simply add the number of choices from each group to get the total number of options.

Property

According to the Multiplication Principle, 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. This is also known as the Fundamental Counting Principle.

Examples

• A restaurant meal special includes one of 4 main courses, one of 3 side dishes, and one of 2 drinks. The total number of different meal combinations is $4 \times 3 \times 2 = 24$.
• To create a user ID, you must choose two letters and then three digits. The number of possible user IDs is $26 \times 26 \times 10 \times 10 \times 10 = 676,000$.
• A student is choosing their schedule, with 5 options for a math class and 4 options for a science class. They can create a schedule with one of each in $5 \times 4 = 20$ ways.

Explanation

This is the 'and' rule for sequential choices. When you have to perform one task AND then another, you multiply the number of options for each task to find the total number of possible outcomes or combinations.

Property

An ordering of objects is called a permutation. Given $n$ distinct objects, the number of ways to select $r$ objects from the set in order is

To solve, first determine the total number of options, $n$. Then determine the number of options being selected, $r$. Finally, substitute these values into the formula and evaluate.

Examples

• In a race with 8 runners, the number of ways to award gold, silver, and bronze medals is $P(8, 3) = \frac{8!}{(8-3)!} = 336$.
• A manager needs to schedule 4 out of 7 employees for four different shifts. The number of ways she can assign them is $P(7, 4) = \frac{7!}{(7-4)!} = 840$.
• A child has 6 different toy cars and wants to line up 3 of them on a shelf. The number of possible arrangements is $P(6, 3) = \frac{6!}{(6-3)!} = 120$.

Explanation

Permutations count arrangements where the order of selection is important. Think of it like awarding 1st, 2nd, and 3rd place ribbons in a race—who gets which ribbon matters. The formula calculates all possible ordered arrangements.

Property

Given $n$ distinct objects, the number of ways to select $r$ objects from the set is

To find the number of combinations, identify the total number of items, $n$, and the number you are choosing, $r$. Then, replace the variables in the formula with the given values and evaluate.

Examples

• An art teacher wants to choose 4 students from her class of 20 to go on a field trip. The number of ways she can choose the group is $C(20, 4) = \frac{20!}{4!(20-4)!} = 4,845$.
• From a deck of 52 cards, the number of different 5-card hands you can be dealt is $C(52, 5) = \frac{52!}{5!(52-5)!} = 2,598,960$.
• A student must answer 5 out of 8 questions on an exam. The number of ways to choose which questions to answer is $C(8, 5) = \frac{8!}{5!(8-5)!} = 56$.

Explanation

Combinations count selections where order does not matter. Think of picking a team from a group of players—it doesn't matter if you were picked first or last, you're still on the team. This formula finds the number of unique groups.

Property

A set containing $n$ distinct objects has $2^n$ subsets. This is because for each of the $n$ objects, we have two choices: either include it in the subset or not. This gives a total of $2 \times 2 \times \dots \times 2$ ($n$ times) possible subsets.

Examples

• A pizza place offers 8 different toppings. The total number of different pizzas that can be made (including a plain pizza) is $2^8 = 256$.
• A survey contains 10 true/false questions. The total number of ways a person can fill out the survey is $2^{10} = 1,024$.
• A company has 6 project add-on features a client can choose. The number of possible feature packages, including choosing none, is $2^6 = 64$.

Explanation

This formula calculates every possible grouping that can be formed from a set of items, including taking no items (the empty set) or all items. For each item, you simply make a 'yes' or 'no' choice to include it.

Property

If there are $n$ elements in a set and $r_1$ are alike, $r_2$ are alike, $r_3$ are alike, and so on through $r_k$, the number of permutations can be found by

Examples

• Find the number of distinct arrangements of the letters in the word 'MISSISSIPPI'. There are 11 letters total, with 4 I's, 4 S's, and 2 P's. The number is $\frac{11!}{4!4!2!} = 34,650$.
• A decorator is arranging 10 balloons in a line. If 5 are red, 3 are blue, and 2 are white, the number of unique arrangements is $\frac{10!}{5!3!2!} = 2,520$.
• How many different ways can you arrange the letters in the word 'STATISTICS'? There are 10 letters, with 3 S's, 3 T's, and 2 I's. The number of ways is $\frac{10!}{3!3!2!} = 50,400$.

Explanation

When arranging items that include duplicates, you must correct for overcounting. This formula finds the number of unique arrangements by dividing the total permutations ($n!$) by the permutations of each group of identical items.