MTH120 College Algebra Chapter 9.5

 9.5 Counting Principles

Counting principles are fundamental concepts in combinatorics and probability theory that help us count and analyze the number of possible outcomes or arrangements in various scenarios. Two key counting principles are the Fundamental Counting Principle and the Multiplication Principle. These principles are fundamental in solving combinatorial problems.

1. Fundamental Counting Principle: This principle is also known as the "Counting Rule" or the "Product Rule." It states that if there are $�$ ways to do one thing and $�$ ways to do another thing, then there are $�\times �$ ways to do both things.

   For example, if you have 3 choices for a shirt and 4 choices for pants, you have $3\times 4=12$ possible outfit combinations.

2. Multiplication Principle: This principle is a generalization of the Fundamental Counting Principle. It states that if there are $�$ different independent events or tasks, each with ${�}_{�}$ possible outcomes, then the total number of outcomes for the sequence of events is ${�}_1\times {�}_2\times \dots \times {�}_{�}$.

   For instance, if you have 4 choices for breakfast, 3 choices for lunch, and 2 choices for dinner, the total number of meal combinations for a day is $4\times 3\times 2=24$.

In addition to these fundamental counting principles, other counting techniques are essential in combinatorics:

3. Permutations: A permutation is an ordered arrangement of objects. Permutations can be used to count the number of ways to arrange $�$ objects out of $�$ distinct objects in a specific order. The formula for permutations is $�(�,�)=\frac{�!}{(�-�)!}$, where $�$ is the total number of objects, $�$ is the number of objects to arrange, and $�!$ represents the factorial of $�$.

4. Combinations: A combination is an unordered selection of objects. Combinations are used to count the number of ways to select $�$ objects from a set of $�$ objects without considering the order. The formula for combinations is $�(�,�)=\frac{�!}{�!(�-�)!}$, where $�$ is the total number of objects, $�$ is the number of objects to select, and $�!$ represents the factorial of $�$.

5. Binomial Coefficients: Binomial coefficients, often denoted as $\left(\genfrac{}{}{0px}{}{�}{�}\right)$, represent the number of ways to choose $�$ items from $�$ items without considering their order. Binomial coefficients are used in binomial expansion and probability calculations.

Counting principles and techniques are crucial in solving problems related to permutations, combinations, probability, and many other areas of mathematics and science. They provide a systematic approach to counting and organizing possibilities.

The Addition Principle, also known as the Addition Rule, is a fundamental counting principle in combinatorics. It is used to count the number of ways an event can occur when there are multiple mutually exclusive ways for it to happen. The principle states that if an event can occur in $�$ ways and another independent event can occur in $�$ ways, then there are $�+�$ ways for either event to occur.

Here are some common scenarios where the Addition Principle is applied:

1. Mutually Exclusive Events: When events are mutually exclusive, meaning that they cannot occur simultaneously, you can use the Addition Principle. For example, when counting the ways to select an apple or an orange from a fruit basket, the events are mutually exclusive. If there are 4 apples and 3 oranges, there are $4+3=7$ ways to make a selection.

2. Overlap in Event Counting: In some cases, events may have some overlap in counting. The Addition Principle can be applied by counting the number of ways for each event and then correcting for the overlap. For instance, when counting the number of people who like both ice cream and cake, you count those who like ice cream and those who like cake, and then subtract those who like both.

3. Multiple Criteria Selection: When you need to select items based on multiple criteria and there is no overlap, the Addition Principle is used. For instance, when choosing a pizza topping from 3 choices (cheese, pepperoni, mushrooms) and a type of crust from 2 choices (thin, thick), there are $3+2=5$ different combinations.

4. Complementary Events: In probability, the Addition Principle is often applied to calculate the probability of the complement of an event. If $�$ is an event, then the probability of either $�$ or its complement $\overline{�}$ happening is $�(�)+�(\overline{�})=1$.

In summary, the Addition Principle is a simple but powerful tool for counting the number of ways events can occur, especially when dealing with mutually exclusive or non-overlapping scenarios. It's widely used in combinatorics and probability theory.

The Multiplication Principle is a fundamental counting principle used to determine the number of possible outcomes in a sequence of independent events. It is also known as the "Counting Principle." The principle states that if there are $�$ ways for one event to occur and $�$ ways for another independent event to occur, then there are $�\times �$ ways for both events to occur together.

Here are some examples of how to use the Multiplication Principle:

Example 1: Arranging Outfits

Suppose you have 3 choices for a shirt (e.g., red, blue, green) and 4 choices for pants (e.g., jeans, khakis, shorts, skirt). How many different outfits can you create by choosing one shirt and one pair of pants?

Using the Multiplication Principle:

• Number of choices for a shirt ($�$) = 3
• Number of choices for pants ($�$) = 4

Total outfits = $�\times �=3\times 4=12$

So, there are 12 different outfit combinations you can create.

Example 2: Passwords

Suppose you want to create a 4-character password using the characters A, B, C, and D. How many different passwords can you create?

Using the Multiplication Principle:

• For each character in the password, there are 4 choices (A, B, C, D).

Total number of passwords = $4\times 4\times 4\times 4=4^4=256$

So, you can create 256 different passwords using these characters.

Example 3: Seating Arrangement

In a classroom, there are 5 students, and you want to find out how many different ways they can be seated in a row of 5 chairs. Assume that the order matters, and students are distinct.

Using the Multiplication Principle:

• For the first seat, there are 5 choices.
• For the second seat, there are 4 choices remaining (since one student is already seated).
• For the third seat, there are 3 choices.
• For the fourth seat, there are 2 choices.
• For the fifth seat, there is 1 choice.

Total number of seating arrangements = $5\times 4\times 3\times 2\times 1=5!=120$

So, there are 120 different seating arrangements for the 5 students.

The Multiplication Principle is a fundamental concept in combinatorics and is useful for counting the total number of outcomes when you have a sequence of independent events. It can be applied to a wide range of scenarios, from arranging objects to creating passwords or solving permutations problems.

To find the number of permutations of $�$ distinct objects, you can use the formula for permutations. A permutation is an ordered arrangement of objects, and the number of permutations of $�$ distinct objects taken $�$ at a time is represented as $�(�,�)$.

The general formula for permutations is:

$�(�,�)=\frac{�!}{(�-�)!}$

Where:

• $�(�,�)$ is the number of permutations of $�$ distinct objects taken $�$ at a time.
• $�!$ represents the factorial of $�$, which is the product of all positive integers from 1 to $�$.
• $(�-�)!$ represents the factorial of $(�-�)$.

To find the number of permutations of $�$ distinct objects, you need to specify how many of those objects you want to arrange ($�$). If you want to arrange all $�$ objects, then $�=�$, and you use the formula as is.

For example, if you have 5 distinct objects and want to find the number of permutations of all 5 objects (i.e., arranging all of them), you would use:

$�(5,5)=\frac{5!}{(5-5)!}=\frac{5!}{0!}=5!$

Calculate $5!$ to find the number of permutations.

However, if you want to arrange only a subset of the $�$ objects (e.g., arranging 3 out of 5 objects), you adjust $�$ accordingly in the formula.

For example, if you want to find the number of permutations of 3 out of 5 distinct objects, you would use:

$�(5,3)=\frac{5!}{(5-3)!}=\frac{5!}{2!}$

Calculate $\frac{5!}{2!}$ to find the number of permutations in this specific scenario.

The formula for permutations is a fundamental concept in combinatorics and is used in various situations where the arrangement of objects matters. It's essential for solving problems involving permutations, arrangements, and ordered selections.

The number of permutations of $�$ distinct objects can also be found using the Multiplication Principle. This principle is a useful way to think about permutations and is especially helpful when you need to find permutations for different positions.

Here's how you can use the Multiplication Principle to find the number of permutations of $�$ distinct objects:

1. Consider the first position for an object. You have $�$ choices for this position.

2. After placing the first object, you have $�-1$ choices for the second position because one object has already been placed.

3. For the third position, you have $�-2$ choices since two objects are already placed.

4. Continue this process until you reach the $�$th and final position, for which you have $1$ choice.

5. To find the total number of permutations, multiply the number of choices for each position.

In mathematical terms, the number of permutations of $�$ distinct objects is $�\times (�-1)\times (�-2)\times \dots \times 2\times 1$, which is equal to $�!$ (read as "n factorial").

So, the number of permutations of $�$ distinct objects is $�!$.

For example, if you have 5 distinct objects, the number of permutations is $5!$, which is:

$5!=5\times 4\times 3\times 2\times 1=120$

So, there are 120 permutations of 5 distinct objects.

Let's find the number of permutations of $�$ distinct objects using the permutation formula $�(�)=�!$. I'll provide some examples:

Example 1: Find the number of permutations of 4 distinct objects.

For $�=4$, you can use the formula $�(�)=�!$:

$�(4)=4!=4\times 3\times 2\times 1=24$

So, there are 24 permutations of 4 distinct objects.

Example 2: Find the number of permutations of 7 distinct objects.

For $�=7$, you can use the formula $�(�)=�!$:

$�(7)=7!=7\times 6\times 5\times 4\times 3\times 2\times 1=5,040$

So, there are 5,040 permutations of 7 distinct objects.

Example 3: Find the number of permutations of 3 distinct objects.

For $�=3$, you can use the formula $�(�)=�!$:

$�(3)=3!=3\times 2\times 1=6$

So, there are 6 permutations of 3 distinct objects.

In each of these examples, we used the formula $�(�)=�!$ to find the number of permutations for the given value of $�$. The factorial ($�!$) represents the product of all positive integers from 1 to $�$

The formula for finding the number of permutations of $�$ distinct objects taken $�$ at a time is as follows:

$�(�,�)=\frac{�!}{(�-�)!}$

Where:

• $�(�,�)$ represents the number of permutations of $�$ distinct objects taken $�$ at a time.
• $�!$ is the factorial of $�$, which is the product of all positive integers from 1 to $�$.
• $(�-�)!$ is the factorial of $�-�$, which is the product of all positive integers from 1 to $�-�$.

However, when you want to find the number of permutations of all $�$ distinct objects (i.e., arranging all of them), the formula simplifies to:

$�(�,�)=�!$

In this case, you don't need to subtract anything from $�$ in the formula because you are arranging all $�$ objects.

For example, if you have 5 distinct objects and want to find the number of permutations of all 5 objects, you can use:

$�(5,5)=5!$

This formula is used to calculate the total number of ways to arrange $�$ distinct objects in a specific order.

The formula for finding the number of combinations of $�$ distinct objects taken $�$ at a time is as follows:

$�(�,�)=\frac{�!}{�!(�-�)!}$

Where:

• $�(�,�)$ represents the number of combinations of $�$ distinct objects taken $�$ at a time.
• $�!$ is the factorial of $�$, which is the product of all positive integers from 1 to $�$.
• $�!$ is the factorial of $�$, which is the product of all positive integers from 1 to $�$.
• $(�-�)!$ is the factorial of $�-�$, which is the product of all positive integers from 1 to $�-�$.

Let's use this formula to find the number of combinations in a couple of examples:

Example 1: Find the number of ways to choose 2 items from a set of 5 distinct items.

For this example, $�=5$ (total number of distinct items) and $�=2$ (number of items to choose). Using the formula:

$�(5,2)=\frac{5!}{2!(5-2)!}=\frac{5!}{2!\cdot 3!}$

Now, calculate the factorials:

$5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$ $2!=2\cdot 1=2$ $3!=3\cdot 2\cdot 1=6$

Now, plug these values into the formula:

$�(5,2)=\frac{120}{2\cdot 6}=\frac{120}{12}=10$

So, there are 10 different combinations of choosing 2 items from a set of 5 distinct items.

Example 2: Find the number of ways to select 3 students out of a class of 25 for a special project.

In this case, $�=25$ (total number of students) and $�=3$ (number of students to select). Using the formula:

$�(25,3)=\frac{25!}{3!(25-3)!}=\frac{25!}{3!\cdot 22!}$

Calculate the factorials:

$25!=\text{(a very large number)}$ $3!=3\cdot 2\cdot 1=6$ $22!=\text{(a very large number)}$

Now, plug these values into the formula:

$�(25,3)=\frac{\text{(a very large number)}}{6\cdot \text{(a very large number)}}$

The result will be a large number, but it represents the number of ways to select 3 students out of 25 for the special project.

The formula for combinations is used in a wide range of scenarios where you need to count the number of ways to choose a subset of items from a larger set without considering the order.

To find the number of subsets of a set with $�$ elements, you can use the formula $2^{�}$. This formula tells you how many subsets, including the empty set and the set itself, can be formed from a set of $�$ elements.

Let's look at some examples:

Example 1: Find the number of subsets of a set with 3 elements.

For a set with 3 elements, $�=3$. Using the formula $2^{�}$:

Number of subsets = $2^3=8$

So, there are 8 subsets: $\{\},\{�\},\{�\},\{�\},\{�,�\},\{�,�\},\{�,�\},\{�,�,�\}$.

Example 2: Find the number of subsets of a set with 4 elements.

For a set with 4 elements, $�=4$. Using the formula $2^{�}$:

Number of subsets = $2^4=16$

So, there are 16 subsets: ({}, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}.

In general, the formula $2^{�}$ is a quick way to find the number of subsets of a set with $�$ elements. It's important to note that this formula includes the empty set and the set itself as subsets.

The number of permutations of $�$ non-distinct objects (objects that are not unique or indistinguishable from each other) can be calculated using the formula:

$$\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">;</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">1</span>}}<span\; style="background-color:\; white;">,</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">,</span><span\; style="background-color:\; white;">\dots </span><span\; style="background-color:\; white;">,</span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">�</span>}}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">!</span>}{{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">1</span>}}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">2</span>}}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span><span\; style="background-color:\; white;">\dots </span><span\; style="background-color:\; white;">\cdot </span>{\mathrm{<span\; style="background-color:\; white;">�</span>}}_{\mathrm{<span\; style="background-color:\; white;">�</span>}}<span\; style="background-color:\; white;">!</span>}$$

Where:

• $�(�;{�}_1,{�}_2,\dots ,{�}_{�})$ is the number of permutations of $�$ non-distinct objects with $�$ distinct groups, where each group has ${�}_{�}$ identical objects.
• $�$ is the total number of non-distinct objects.
• ${�}_1,{�}_2,\dots ,{�}_{�}$ are the counts of objects in each distinct group.
• $�!$ is the factorial of $�$, which is the product of all positive integers from 1 to $�$.
• ${�}_{�}!$ is the factorial of ${�}_{�}$, which is the product of all positive integers from 1 to ${�}_{�}$.

Here are some examples to illustrate how to find the number of permutations of non-distinct objects:

Example 1: Find the number of permutations of the letters in the word "BOOKKEEPER."

In this word, there are:

• 3 "E"s,
• 2 "O"s,
• 2 "K"s, and
• 1 "B", 1 "P", and 1 "R."

So, we have $�=9$ non-distinct objects ($�$ is the total count of letters), and we have several groups of distinct letters: {E, O, K, B, P, R}.

Using the formula:

$$\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">9</span>}<span\; style="background-color:\; white;">;</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">9</span>}<span\; style="background-color:\; white;">!</span>}{\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span>}$$

Calculate the factorials:

$$\mathrm{<span\; style="background-color:\; white;">9</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">9</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">8</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">5</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">4</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">362</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">880</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">6</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">2</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">1</span>}$$

Now, plug these values into the formula:

$$\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">9</span>}<span\; style="background-color:\; white;">;</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">362</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">880</span>}}{\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">362</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">880</span>}\mathrm{<span\; style="background-color:\; white;">/</span>}\mathrm{<span\; style="background-color:\; white;">24</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">15</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">120</span>}$$

So, there are 15,120 permutations of the letters in the word "BOOKKEEPER."

Example 2: Find the number of permutations of the letters in the word "SUCCESS."

In this word, there are:

• 3 "S"s,
• 2 "C"s, and
• 1 "E" and 1 "U."

So, we have $�=7$ non-distinct objects ($�$ is the total count of letters), and we have distinct groups of letters: {S, C, E, U}.

Using the formula:

$$\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">;</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">!</span>}{\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span>}$$

Calculate the factorials:

$$\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">5</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">4</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">5</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">040</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">6</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">2</span>}$$

$$\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">!</span><span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">1</span>}$$

Now, plug these values into the formula:

$$\mathrm{<span\; style="background-color:\; white;">�</span>}<span\; style="background-color:\; white;">(</span>\mathrm{<span\; style="background-color:\; white;">7</span>}<span\; style="background-color:\; white;">;</span>\mathrm{<span\; style="background-color:\; white;">3</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">)</span><span\; style="background-color:\; white;">=</span>\frac{\mathrm{<span\; style="background-color:\; white;">5</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">040</span>}}{\mathrm{<span\; style="background-color:\; white;">6</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">2</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}<span\; style="background-color:\; white;">\cdot </span>\mathrm{<span\; style="background-color:\; white;">1</span>}}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">5</span>}<span\; style="background-color:\; white;">,</span>\mathrm{<span\; style="background-color:\; white;">040</span>}\mathrm{<span\; style="background-color:\; white;">/</span>}\mathrm{<span\; style="background-color:\; white;">12</span>}<span\; style="background-color:\; white;">=</span>\mathrm{<span\; style="background-color:\; white;">420</span>}$$

So, there are 420 permutations of the letters in the word "SUCCESS."

In both examples, the formula for permutations of non-distinct objects was used to account for the presence of identical letters.