Step 1: Analyzing the Positional Constraints

We are required to form a 3-digit number using the set of digits $S = \{1, 2, 3, 4, 5\}$. A 3-digit number consists of three distinct positions: the Hundreds place, the Tens place, and the Units place.

Step 2: Applying the Fundamental Principle of Counting (Multiplication Principle)

Since the repetition of digits is allowed, the selection of a digit for one position does not restrict the choices for the subsequent positions. According to the Fundamental Principle of Counting, if an operation can be performed in $n$ ways, and a subsequent independent operation can be performed in $m$ ways, then the total number of ways to perform both is $n \times m$.

For a 3-digit number:

• Number of ways to fill the Hundreds place ($n_1$): $5$
• Number of ways to fill the Tens place ($n_2$): $5$
• Number of ways to fill the Units place ($n_3$): $5$

Step 3: Mathematical Calculation

The total number of 3-digit numbers is given by the product of the number of choices for each position:

$Total = n_1 \times n_2 \times n_3$

$Total = 5 \times 5 \times 5$

$Total = 5^3$

$Total = 125$

Final Solution: The total number of 3-digit numbers that can be formed is 125.

Step 1: Analyzing the Sample Space for a Single Event

Each individual toss of a coin is an independent experiment. A fair coin has two possible outcomes for each toss, defined as the set $S = \{H, T\}$, where $H$ denotes 'Heads' and $T$ denotes 'Tails'.

Thus, the number of possible outcomes for a single toss, $n_1 = 2$.

Step 2: Applying the Fundamental Counting Principle

The Fundamental Counting Principle (also known as the Multiplication Principle) states that if there are $n$ ways to perform one task and $m$ ways to perform a second task, the total number of ways to perform both is $n \times m$. [Per the Fundamental Counting Principle of Combinatorics].

For $k$ independent events, the total number of outcomes is given by the product of the number of outcomes for each event: $N = n_1 \times n_2 \times n_3 \times \dots \times n_k$.

Step 3: Calculating Total Outcomes for Three Tosses

Given the coin is tossed $k = 3$ times:

• Number of outcomes for toss 1 ($n_1$) = 2
• Number of outcomes for toss 2 ($n_2$) = 2
• Number of outcomes for toss 3 ($n_3$) = 2

Total possible outcomes = $2 \times 2 \times 2 = 2^3$.

Total possible outcomes = $8$.

Visual Representation of the Sample Space

Step 4: Enumerating the Sample Space ($S$)

To verify the calculation, we list the Cartesian product of the three sets $\{H, T\} \times \{H, T\} \times \{H, T\}$:

Final Solution: The total number of possible outcomes when a coin is tossed 3 times is $2^3 = 8$.

Given Variables & Initial Setup

The problem involves the selection and arrangement of flags to generate a signal. We are provided with the following parameters:

• Total number of distinct flags available ($n$): $5$
• Number of flags required to form one signal ($r$): $2$
• Constraint: The signal depends on the order of the flags (one below the other).

[By the Fundamental Principle of Counting, if an operation can be performed in $n$ ways and a second operation can be performed in $m$ ways, then the total number of ways to perform both operations in sequence is $n \times m$.]

Step 1: Analyzing the Selection Process

Since the signal requires two flags placed one below the other, the position matters (top position and bottom position). This is a problem of Permutation, where we calculate the arrangement of $r$ objects out of a set of $n$ distinct objects.

The number of ways to fill the positions is as follows:

• Top Position: Any of the $5$ flags can be placed here. Number of ways = $5$.
• Bottom Position: Since one flag is already used for the top position and the flags must be different, we are left with $(5 - 1) = 4$ flags. Number of ways = $4$.

Step 2: Applying the Fundamental Counting Principle

To find the total number of distinct signals ($N$), we multiply the possibilities for each position:

$N = 5 \times 4$

$N = 20$

Step 3: Verification via Permutation Formula

Alternatively, using the Permutation formula $_nP_r = \frac{n!}{(n-r)!}$:

$_5P_2 = \frac{5!}{(5-2)!}$ [Definition of Permutation]

$_5P_2 = \frac{5 \times 4 \times 3!}{3!}$

$_5P_2 = 5 \times 4 = 20$

Visual Representation of Signal Generation

Final Solution: The total number of different signals that can be generated is 20.

Given Variables & Initial Setup

To determine the total number of 3-digit numbers that can be formed using the set of digits $S = \{1, 2, 3, 4, 5\}$, where repetition is not allowed, we define the parameters:

• Total number of available digits ($n$): $5$
• Number of positions to be filled ($r$): $3$
• Constraint: Repetition is strictly forbidden.

Step 1: Application of the Fundamental Counting Principle

A 3-digit number consists of three distinct positions: the hundreds place, the tens place, and the units place. We analyze the number of choices for each position sequentially:

Step 2: Mathematical Calculation

By the Fundamental Counting Principle (also known as the Multiplication Principle), if one event can occur in $m$ ways and a second independent event can occur in $n$ ways, then both events can occur in $m \times n$ ways. [Per the Fundamental Counting Principle].

Total numbers = $5 \times 4 \times 3$

Total numbers = $60$

Step 3: Verification via Permutation Formula

The problem can also be expressed as finding the number of permutations of $n$ distinct objects taken $r$ at a time, denoted by $^nP_r$.

$^nP_r = \frac{n!}{(n-r)!}$

Substituting $n=5$ and $r=3$:

$^5P_3 = \frac{5!}{(5-3)!} = \frac{5!}{2!}$

$^5P_3 = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60$ [By the definition of Permutations].

Final Solution: 60

Step 1: Identification of Parameters

The problem asks for the number of arrangements (codes) of 4 letters chosen from a set of 10 distinct letters. Since the order of letters in a code matters (e.g., $ABCD \neq DCBA$), this is a permutation problem. [Per the Fundamental Principle of Counting regarding ordered arrangements].

• Total number of available letters ($n$) = $10$
• Number of positions to fill ($r$) = $4$
• Constraint: No letter can be repeated (Sampling without replacement).

Step 2: Mathematical Formulation

The number of permutations of $n$ distinct objects taken $r$ at a time is given by the formula:

$P(n, r) = \frac{n!}{(n - r)!}$

Substituting the given values $n = 10$ and $r = 4$:

$P(10, 4) = \frac{10!}{(10 - 4)!}$

$P(10, 4) = \frac{10!}{6!}$

Step 3: Algebraic Calculation

Expanding the factorial terms to simplify the expression:

$P(10, 4) = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!}$

Canceling the $6!$ term from the numerator and denominator:

$P(10, 4) = 10 \times 9 \times 8 \times 7$

$10 \times 9 = 90$

$90 \times 8 = 720$

$720 \times 7 = 5040$

Step 4: Logical Verification via Multiplication Principle

Alternatively, by the Multiplication Principle [Per the Fundamental Counting Principle], we have 4 slots to fill:

Total combinations = $10 \times 9 \times 8 \times 7 = 5040$.

Final Solution: The total number of 4-letter codes that can be formed is 5040.

Step 1: Analyzing the Structural Constraints

To form a $3$-digit number, we have three positions to fill: the Hundreds place, the Tens place, and the Units (ones) place. Let the positions be represented by $H$, $T$, and $U$ respectively.

• The available digits are the set $S = \{1, 2, 3, 4, 5, 6\}$.
• Total number of available digits $n = 6$.
• Constraint 1: The number must be even. [By definition, a number is even if its unit digit is divisible by $2$].
• Constraint 2: Repetition of digits is allowed.

Step 2: Determining Permutations for the Units Place ($U$)

For the number to be even, the digit in the units place must be chosen from the set of even digits available in $S$. The even digits in $S$ are $\{2, 4, 6\}$.

Number of choices for the units place ($n_U$) = $3$.

Step 3: Determining Permutations for Remaining Places ($H$ and $T$)

Since repetition is permitted [Per the Fundamental Counting Principle], the choice of one position does not restrict the choices for the others.

• For the Hundreds place ($H$): We can select any of the $6$ digits from $S$. Thus, $n_H = 6$.
• For the Tens place ($T$): We can select any of the $6$ digits from $S$. Thus, $n_T = 6$.

Step 4: Calculation via Fundamental Counting Principle

The total number of ways to form the $3$-digit even number is the product of the number of ways to fill each position [By the Multiplication Principle of Counting]:

$\text{Total Numbers} = n_H \times n_T \times n_U$

$\text{Total Numbers} = 6 \times 6 \times 3$

$\text{Total Numbers} = 36 \times 3 = 108$

Visual Representation of the Choice Tree

Final Solution: 108 even numbers

Given Variables & Initial Setup

A 5-digit telephone number is to be constructed using the set of available digits $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$. The total number of available digits $n = 10$.

• Total positions available: $5$
• Fixed starting digits: $67$ (occupying the first two positions).
• Constraint 1: The first two positions are occupied by $6$ and $7$.
• Constraint 2: No digit can be repeated (sampling without replacement).

Step 1: Identifying Remaining Positions and Digits

Since the first two positions are fixed as $67$, we are required to fill the remaining $3$ positions ($5 - 2 = 3$).

The set of available digits for the remaining positions is $S' = S \setminus \{6, 7\} = \{0, 1, 2, 3, 4, 5, 8, 9\}$.

The number of available digits remaining is $n' = 10 - 2 = 8$.

Step 2: Applying the Fundamental Principle of Counting

We must select and arrange $3$ digits from the set of $8$ available digits to fill the $3$ vacant spots. Per the [Fundamental Counting Principle], if an event can occur in $m$ ways and a subsequent event in $n$ ways, the total occurrences are $m \times n$.

Step 3: Calculating Permutations

The number of ways to arrange $r$ objects from $n$ distinct objects is given by the formula $P(n, r) = \frac{n!}{(n-r)!}$.

Here, $n = 8$ and $r = 3$:

$P(8, 3) = \frac{8!}{(8-3)!}$

$P(8, 3) = \frac{8 \times 7 \times 6 \times 5!}{5!}$ [By the definition of factorial simplification]

$P(8, 3) = 8 \times 7 \times 6 = 336$

Final Solution: The total number of 5-digit telephone numbers that can be constructed is 336.