Step 1: Theoretical Foundation of Permutations

The permutation formula for $n$ distinct objects taken $r$ at a time is defined as: $${}^n P_r = \frac{n!}{(n-r)!}$$ [Per the Fundamental Principle of Permutations where $0 \leq r \leq n$].

Step 2: Expanding the Given Equation

Given the equation ${}^5 P_r = 2 \cdot {}^6 P_{r-1}$ (Assuming standard textbook context for this specific problem type, where the factor 2 is often implied in this specific exercise; however, strictly following the provided prompt ${}^5 P_r = {}^6 P_{r-1}$):

Applying the formula to both sides:

$$\frac{5!}{(5-r)!} = \frac{6!}{(6-(r-1))!}$$

Simplifying the denominator on the right side:

$$\frac{5!}{(5-r)!} = \frac{6!}{(7-r)!}$$

Step 3: Algebraic Manipulation

To solve for $r$, we expand the factorials to isolate common terms:

$$\frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r) \times (6-r) \times (5-r)!}$$

[By the property of factorials: $n! = n \times (n-1)!$]

Dividing both sides by $5!$ and multiplying by $(5-r)!$ (assuming $5-r \geq 0$):

$$1 = \frac{6}{(7-r)(6-r)}$$

Step 4: Solving the Quadratic Equation

Cross-multiplying yields:

$$(7-r)(6-r) = 6$$

Expanding the binomials:

$$42 - 7r - 6r + r^2 = 6$$

$$r^2 - 13r + 42 = 6$$

$$r^2 - 13r + 36 = 0$$

Factoring the quadratic equation:

$$(r - 9)(r - 4) = 0$$

Step 5: Applying Domain Constraints

The solutions are $r = 9$ or $r = 4$.

• For ${}^5 P_r$, the condition $r \leq 5$ must hold.
• If $r = 9$, the term ${}^5 P_9$ is undefined as $n < r$.
• If $r = 4$, the conditions $4 \leq 5$ and $3 \leq 6$ are satisfied.

[By the definition of Permutations, $n \geq r$ is a strict requirement].

Final Solution: r = 4

Step 1: Analyzing the Word Composition

The given word is PERMUTATIONS. We perform an exhaustive count of its constituent letters:

• Total letters ($n$) = 12
• Letter frequencies:
  • P: 1, E: 1, R: 1, M: 1, U: 1, T: 2, A: 1, I: 1, O: 1, N: 1, S: 1
• Total distinct letters: 11 distinct types, with 'T' repeating twice.

Step 2: Segregating Vowels and Consonants

To arrange the letters such that all vowels are kept together, we identify the specific vowels in the word:

• Vowels: {E, U, A, I, O} (Total = 5)
• Consonants: {P, R, M, T, T, N, S} (Total = 7)

[By the method of grouping, we treat the set of vowels as a single entity or 'super-letter' (V)].

Step 3: Calculating Internal Permutations of the Vowel Group

The 5 vowels {E, U, A, I, O} are all distinct. The number of ways to arrange these 5 vowels internally is given by $5!$ [Per the Multiplication Principle of counting]:

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

Step 4: Arranging the Combined Entities

After grouping the vowels into a single unit (V), we now have the following entities to arrange:

• Consonants: {P, R, M, T, T, N, S} (7 entities)
• Vowel group: {V} (1 entity)
• Total entities to arrange = $7 + 1 = 8$

Among these 8 entities, the letter 'T' appears twice. Therefore, the number of distinct permutations of these 8 entities is calculated as:

$\frac{8!}{2!}$

Calculation: $\frac{40320}{2} = 20160$

Step 5: Final Combinatorial Synthesis

The total number of arrangements is the product of the external arrangements of the entities and the internal arrangements of the vowels [Per the Fundamental Counting Principle]:

$\text{Total} = (\text{Arrangements of entities}) \times (\text{Internal vowel permutations})$

$\text{Total} = 20160 \times 120$

$\text{Total} = 2,419,200$

Visual Representation of the Grouping Method

Final Solution: 2,419,200 ways

Step 1: Analyzing the Word and Identifying Constraints

The given word is PERMUTATIONS. First, we identify the composition of the word:

• Total number of letters: $12$
• Frequency of letters:

  Letter	Frequency
  P	1
  E	1
  R	1
  M	1
  U	1
  T	2
  A	1
  I	1
  O	1
  N	1
  S	1

Constraint: The word must start with P and end with S.

Step 2: Determining Remaining Positions

By fixing P at the first position and S at the last position, we have accounted for $2$ letters. The remaining letters to be arranged in the middle $10$ positions are $\{E, R, M, U, T, T, A, I, O, N\}$.

[By the Fundamental Counting Principle, the total number of arrangements is the product of independent choices.]

Step 3: Calculating Permutations of the Remaining Letters

The number of items to arrange is $n = 10$. Within these $10$ letters, the letter T repeats $2$ times. [Per the formula for permutations of multiset: $P = \frac{n!}{n_1! n_2! ... n_k!}$].

Number of arrangements = $\frac{10!}{2!}$

Step 4: Algebraic Evaluation

We expand the factorial expression:

$\frac{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$

$\frac{3628800}{2} = 1814400$

Final Solution: The total number of ways the letters can be arranged such that the word starts with P and ends with S is $1,814,400$.

Step 1: Analyzing the Word Composition

The word MISSISSIPPI consists of $11$ letters. We must first identify the frequency of each distinct letter to determine the total number of permutations [Per the Permutation Formula for Multisets: $P = \frac{n!}{n_1! n_2! ... n_k!}$].

• Total letters ($n$) = $11$
• $M$: $1$
• $I$: $4$
• $S$: $4$
• $P$: $2$

Step 2: Calculating Total Permutations

The total number of distinct arrangements of the letters in MISSISSIPPI is given by:

Total = $\frac{11!}{1! \times 4! \times 4! \times 2!}$

Total = $\frac{39,916,800}{1 \times 24 \times 24 \times 2} = \frac{39,916,800}{1152} = 34,650$

Step 3: Calculating Permutations where the four I's come together

To find the permutations where all four $I$'s are together, we treat the block of $(IIII)$ as a single entity or "super-letter".

Now, our set of items to arrange becomes $\{M, (IIII), S, S, S, S, P, P\}$.

Number of entities = $1 + 1 + 4 + 2 = 8$ units.

The number of arrangements where $I$'s are together = $\frac{8!}{1! \times 1! \times 4! \times 2!}$

Calculation: $\frac{40,320}{24 \times 2} = \frac{40,320}{48} = 840$

Step 4: Applying the Complementary Counting Principle

Per the Principle of Inclusion-Exclusion, the number of permutations where the four $I$'s do not come together is found by subtracting the cases where they do come together from the total number of permutations.

Formula: $N(\text{not together}) = N(\text{total}) - N(\text{together})$

$N = 34,650 - 840$

$N = 33,810$

Visual Representation of the Permutation Constraint

Final Solution: 33,810

Step 1: Identifying the Set of Letters

The word given is MONDAY. We must first identify the constituent letters to determine the total count and the subsets of vowels and consonants.

• Total letters: $M, O, N, D, A, Y$ (Total = $6$ unique letters).
• Vowels: $O, A$ (Total = $2$).
• Consonants: $M, N, D, Y$ (Total = $4$).

Step 2: Defining the Positional Constraints

We are forming a $6$-letter word using all letters of MONDAY without repetition, subject to the condition that the first letter must be a vowel.

Let the positions be represented as $P_1, P_2, P_3, P_4, P_5, P_6$.

Constraint: $P_1 \in \{O, A\}$.

Step 3: Calculating Permutations for the Restricted Position

Since the first position ($P_1$) must be filled by a vowel, and there are $2$ available vowels ($O$ or $A$), the number of ways to fill the first position is:

$N(P_1) = 2$ [By Fundamental Counting Principle: Multiplication Rule].

Step 4: Calculating Permutations for the Remaining Positions

After selecting $1$ vowel for the first position, we are left with $5$ remaining letters to be arranged in the remaining $5$ positions ($P_2, P_3, P_4, P_5, P_6$).

The number of ways to arrange $5$ distinct items in $5$ spots is given by $5!$ [Per the permutation formula $^nP_r = \frac{n!}{(n-r)!}$ where $n=5, r=5$].

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

Step 5: Final Computation

Total number of words = (Ways to choose $P_1$) $\times$ (Ways to arrange remaining letters).

Total = $2 \times 120 = 240$.

Final Solution: 240 words

Step 1: Identifying the Mathematical Framework

The problem requires the selection of two distinct positions—Chairman and Vice Chairman—from a set of $n = 8$ distinct individuals. Because the roles are distinct (the order of selection matters; choosing Person A as Chairman and Person B as Vice Chairman is different from choosing Person B as Chairman and Person A as Vice Chairman), this is a Permutation problem.

Mathematically, we are calculating the number of arrangements of $r=2$ objects chosen from a total of $n=8$ objects [Per the Fundamental Counting Principle and Permutation definition].

Step 2: Mathematical Formulation

The number of ways to arrange $r$ items out of $n$ distinct items is given by the formula:

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

Where:

• $n = 8$ (Total number of persons available)
• $r = 2$ (Number of positions to be filled)

Step 3: Execution of Calculation

Substitute the given variables into the formula:

${}^8P_2 = \frac{8!}{(8-2)!} = \frac{8!}{6!}$

Expand the factorial terms to simplify the expression [By the definition of factorial $n! = n \times (n-1) \times (n-2) \times \dots \times 1$]:

${}^8P_2 = \frac{8 \times 7 \times 6!}{6!}$

Cancel out the common $6!$ term:

${}^8P_2 = 8 \times 7 = 56$

Visual Representation of Selection Logic

Step 4: Logical Verification via Multiplication Principle

Alternatively, using the Fundamental Principle of Counting:

• Step A: Selection of a Chairman. Since there are 8 persons, there are 8 ways to fill this position.
• Step B: Selection of a Vice Chairman. Since one person cannot hold two positions, there are $(8 - 1) = 7$ remaining candidates.

By the Multiplication Principle [Per the Fundamental Counting Principle], the total number of ways to perform both steps is $8 \times 7 = 56$.

Final Solution: The total number of ways to choose a chairman and a vice chairman is 56.

Step 1: Analyis of the Given Word

The word "PERMUTATIONS" consists of $12$ letters. Let us identify the frequency of each letter:

• P: 1, E: 1, R: 1, M: 1, U: 1, T: 2, A: 1, I: 1, O: 1, N: 1, S: 1.
• Total letters ($n$) = 12.
• Repeating letters: 'T' appears 2 times.

Step 2: Defining the Constraint

We are required to arrange the letters such that there are exactly 4 letters between 'P' and 'S'.

Let the positions be represented by indices $1, 2, 3, \dots, 12$. If P is at position $i$, then S must be at position $i+5$ (to allow for 4 letters between them) or vice versa. The possible pairs of positions $(P, S)$ are:

$(1, 6), (2, 7), (3, 8), (4, 9), (5, 10), (6, 11), (7, 12)$.

There are $7$ such pairs. Since P and S can be swapped, there are $7 \times 2 = 14$ ways to place P and S.

Step 3: Calculating Permutations of Remaining Letters

Once P and S are fixed in a chosen pair of positions, there are $12 - 2 = 10$ remaining positions to be filled by the remaining $10$ letters.

The remaining letters are $\{E, R, M, U, T, T, A, I, O, N\}$.

The number of ways to arrange these 10 letters, where 'T' repeats 2 times, is given by the formula for permutations of multisets [Per Fundamental Principle of Counting]:

$$\text{Ways} = \frac{10!}{2!}$$

Calculating the value:

$$\frac{3,628,800}{2} = 1,814,400$$

Step 4: Final Computation

The total number of arrangements is the product of the number of ways to place P and S and the number of ways to arrange the remaining letters.

$$\text{Total Arrangements} = (\text{Ways to place P and S}) \times (\text{Ways to arrange remaining 10 letters})$$

$$\text{Total Arrangements} = 14 \times 1,814,400$$

$$\text{Total Arrangements} = 25,401,600$$

Visual Representation of Position Slots

Final Solution: The total number of ways to arrange the letters such that there are always 4 letters between P and S is 25,401,600.

Step 1: Expanding the Permutation Formula

The formula for permutations of $n$ objects taken $r$ at a time is defined as:

${}^nP_r = \frac{n!}{(n-r)!}$ [Per the Fundamental Definition of Permutations]

We are given the ratio ${}^{n-1}P_3 : {}^nP_4 = 1 : 9$. Expressing this in fractional form:

$\frac{{}^{n-1}P_3}{{}^nP_4} = \frac{1}{9}$

Step 2: Applying the Permutation Formula to Given Terms

Substitute the specific values into the formula:

${}^{n-1}P_3 = \frac{(n-1)!}{(n-1-3)!} = \frac{(n-1)!}{(n-4)!}$

${}^nP_4 = \frac{n!}{(n-4)!}$

Step 3: Setting up the Algebraic Equation

Divide the two expressions:

$\frac{\frac{(n-1)!}{(n-4)!}}{\frac{n!}{(n-4)!}} = \frac{1}{9}$

By the algebraic property of complex fractions, since the denominators $(n-4)!$ are equivalent, they cancel out:

$\frac{(n-1)!}{n!} = \frac{1}{9}$

Step 4: Simplifying the Factorials

Recalling that $n! = n \times (n-1)!$ [By the Definition of Factorials], we substitute this into the denominator:

$\frac{(n-1)!}{n \times (n-1)!} = \frac{1}{9}$

Canceling the common term $(n-1)!$ from the numerator and denominator:

$\frac{1}{n} = \frac{1}{9}$

Step 5: Solving for $n$

By equating the reciprocal terms:

$n = 9$

Verification of Constraints

For the permutation ${}^nP_r$ to be defined, $n \geq r$.

• For ${}^{n-1}P_3$: $n-1 \geq 3 \implies n \geq 4$
• For ${}^nP_4$: $n \geq 4$

Since $n=9$ satisfies $n \geq 4$, the solution is valid within the domain of the permutation function.

Final Solution: $n = 9$

Step 1: Calculating Total 4-Digit Numbers

To find the total number of 4-digit numbers that can be formed using the digits $\{1, 2, 3, 4, 5\}$ without repetition, we use the Fundamental Counting Principle [Permutation formula $^nP_r = \frac{n!}{(n-r)!}$].

• Total available digits ($n$): 5
• Digits to be selected ($r$): 4

The number of arrangements is given by:

$^5P_4 = \frac{5!}{(5-4)!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{1} = 120$

Step 2: Calculating Even 4-Digit Numbers

A number is even if its unit digit is an even number. From the set $\{1, 2, 3, 4, 5\}$, the even digits are $\{2, 4\}$.

We analyze the 4 positions: _ _ _ _ (Thousands, Hundreds, Tens, Units).

Sub-step 2.1: Fixing the Units Place

To ensure the number is even, the units place must be filled by either $2$ or $4$.

• Number of choices for the units place: $2$ (either 2 or 4).

Sub-step 2.2: Filling the Remaining Places

After fixing one digit in the units place, we have 4 digits remaining from the original set of 5. We must fill the remaining 3 positions (Thousands, Hundreds, Tens).

Number of ways to fill the remaining 3 places = $^4P_3$

$^4P_3 = \frac{4!}{(4-3)!} = \frac{4 \times 3 \times 2 \times 1}{1} = 24$

Step 3: Total Even Permutations

Per the Multiplication Principle of counting, the total count of even numbers is the product of the choices for the units place and the arrangements of the remaining places:

$\text{Total Even Numbers} = (\text{Choices for Units}) \times (\text{Permutations of remaining 3 digits})$

$\text{Total Even Numbers} = 2 \times 24 = 48$

Visual Representation of Logic

Final Solution: The total number of 4-digit numbers is 120, and the number of even 4-digit numbers is 48.

Step 1: Theoretical Foundation

The permutation formula is defined as ${}^nP_r = \frac{n!}{(n-r)!}$, where $0 \leq r \leq n$ [Per the fundamental principle of permutations].

Step 2: Substitution into the Given Equation

Given equation: ${}^5P_r = 2 \cdot {}^6P_{r-1}$

Expanding both sides using the permutation formula:

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(6-(r-1))!}$

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6!}{(7-r)!}$

Step 3: Algebraic Simplification

Expand the factorial terms to identify common factors:

$\frac{5!}{(5-r)!} = 2 \cdot \frac{6 \cdot 5!}{(7-r)(6-r)(5-r)!}$

Cancel $5!$ from both sides and $(5-r)!$ from the denominators:

$1 = \frac{2 \cdot 6}{(7-r)(6-r)}$

$1 = \frac{12}{(7-r)(6-r)}$

Step 4: Solving the Quadratic Equation

Rearrange to form a standard quadratic equation:

$(7-r)(6-r) = 12$

$42 - 7r - 6r + r^2 = 12$

$r^2 - 13r + 42 - 12 = 0$

$r^2 - 13r + 30 = 0$

Factorize the quadratic expression:

$(r - 10)(r - 3) = 0$

Step 5: Verification of Constraints

The roots are $r = 10$ and $r = 3$.

• If $r = 10$: The term ${}^5P_{10}$ is undefined because $n < r$ ($5 < 10$). [Per the domain constraint $r \leq n$].
• If $r = 3$: Check feasibility: ${}^5P_3 = \frac{5!}{2!} = 60$. Right side: $2 \cdot {}^6P_{3-1} = 2 \cdot {}^6P_2 = 2 \cdot \frac{6!}{4!} = 2 \cdot (6 \cdot 5) = 60$.

The condition $60 = 60$ holds true.

Final Solution: r = 3

Given Variables & Initial Setup

The target word is MONDAY.

• Total number of letters ($n$) = 6 (M, O, N, D, A, Y)
• All letters are distinct, meaning there are no repetitions within the set.
• Requirement: Use all letters ($r = 6$) at a time.
• Constraint: No letter is repeated in the formed words.

Step 1: Selecting the Mathematical Principle

To arrange $n$ distinct objects taken $r$ at a time, we utilize the Fundamental Principle of Permutation. [Per the definition of Permutations, the number of arrangements of $n$ distinct objects is given by $n!$]. Since we are using all 6 letters without replacement, we are calculating the total number of permutations of 6 distinct items.

Step 2: Formulating the Calculation

The formula for permutations of $n$ objects taken all at once is: $$P(n, n) = n!$$ Substituting the value $n = 6$: $$P(6, 6) = 6!$$

Step 3: Execution of Factorial Arithmetic

The factorial of 6 is defined as the product of all positive integers from 1 to 6: $$6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1$$ Calculating sequentially:

• $6 \times 5 = 30$
• $30 \times 4 = 120$
• $120 \times 3 = 360$
• $360 \times 2 = 720$
• $720 \times 1 = 720$

Visual Representation of Permutation Slots

Final Solution: The total number of words that can be formed using all letters of the word MONDAY without repetition is 720.

Step 1: Establishing the Mathematical Framework

To determine the number of 3-digit numbers that can be formed using the set of digits $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ without repetition, we employ the Fundamental Principle of Counting, specifically the Permutation formula. Since the order of digits is essential for forming unique numbers, we are selecting and arranging $r$ objects from a set of $n$ distinct objects.

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

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

[Per the Fundamental Counting Principle and the definition of Permutations]

Step 2: Identifying Given Variables

• Total number of distinct digits available ($n$): $9$
• Number of positions to fill (3-digit number, so $r$): $3$
• Constraint: No digit is repeated (sampling without replacement).

Step 3: Execution of the Permutation Calculation

Substituting the values into the formula:

$P(9, 3) = \frac{9!}{(9-3)!}$

$P(9, 3) = \frac{9!}{6!}$

Expanding the factorial terms to simplify:

$P(9, 3) = \frac{9 \times 8 \times 7 \times 6!}{6!}$

$P(9, 3) = 9 \times 8 \times 7$

$P(9, 3) = 72 \times 7 = 504$

Step 4: Logical Verification via Multiplication Principle

Alternatively, using the "slot method" for independent events:

Total arrangements = $9 \times 8 \times 7 = 504$. [By the Multiplication Principle of Counting]

Final Solution: The total number of 3-digit numbers that can be formed is 504.

Step 1: Analyzing the Given Word

The word provided is $MONDAY$.

• The total number of letters in the word is $n = 6$.
• The distinct letters are: $\{M, O, N, D, A, Y\}$. Since all letters are distinct, we have $n = 6$ unique objects to arrange.

Step 2: Defining the Constraint

The problem requires us to form words of $r = 4$ letters at a time, without repetition of letters. [Per the Fundamental Principle of Counting, the number of ways to arrange $r$ objects from $n$ distinct objects is given by the permutation formula $P(n, r)$].

Step 3: Applying the Permutation Formula

The formula for permutations of $n$ distinct objects taken $r$ at a time is defined as:
$P(n, r) = \frac{n!}{(n-r)!}$
Substitute the values $n = 6$ and $r = 4$:
$P(6, 4) = \frac{6!}{(6-4)!}$

Step 4: Algebraic Calculation

Expand the factorials to perform the division:
$P(6, 4) = \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 1}$
Cancel the common terms in the numerator and denominator:
$P(6, 4) = 6 \times 5 \times 4 \times 3$
$P(6, 4) = 30 \times 12$
$P(6, 4) = 360$

Final Solution: The total number of words that can be formed is 360.

Step 1: Identifying the Constraints

To form a 4-digit number using the set of available digits $S = \{0, 1, 2, 3, 4, 5, 6, 7, 8, 9\}$, we must satisfy the following conditions:

• The number consists of 4 distinct positions: Thousands ($T$), Hundreds ($H$), Tens ($t$), and Units ($U$).
• No digit can be repeated [Constraint: Selection without replacement].
• The Thousands place ($T$) cannot be occupied by the digit $0$, as a number starting with $0$ is considered a 3-digit number.

Step 2: Calculating Possibilities for Each Position

We analyze the number of choices for each slot sequentially:

Step 3: Applying the Fundamental Principle of Counting

According to the Fundamental Principle of Multiplication, if one task can be performed in $m$ ways and a second task in $n$ ways, the sequence of tasks can be performed in $m \times n$ ways [Fundamental Counting Principle].

The total number of 4-digit numbers $N$ is calculated as:

$N = (\text{Choices for } T) \times (\text{Choices for } H) \times (\text{Choices for } t) \times (\text{Choices for } U)$

$N = 9 \times 9 \times 8 \times 7$

Step 4: Final Computation

$9 \times 9 = 81$

$81 \times 8 = 648$

$648 \times 7 = 4536$

Final Solution: The total number of 4-digit numbers with no digit repeated is $4536$.

Step 1: Analyzing the Set of Letters

The given word is "EQUATION". To determine the number of permutations, we must first identify the individual components (letters) of the word and check for repetitions.

• Letters: E, Q, U, A, T, I, O, N
• Total number of letters ($n$): 8
• Frequency of letters: Each letter (E, Q, U, A, T, I, O, N) appears exactly once. [Per the Fundamental Counting Principle for distinct objects]

Step 2: Theoretical Foundation

The problem asks for the number of arrangements of $n$ distinct objects taken all at a time. The formula for the number of permutations of $n$ distinct objects is given by $n!$ (n-factorial), where $n! = n \times (n-1) \times (n-2) \times \dots \times 1$. [By the Theorem of Permutations of $n$ distinct objects]

Step 3: Algebraic Calculation

Substituting $n = 8$ into the permutation formula:

Number of arrangements = $8!$

$8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$

Step 4: Visual Representation of Permutation Slots

Final Solution: The total number of words that can be formed is 40,320.

Step 1: Identifying Constraints and Set Parameters

The objective is to form 3-digit even numbers using the set $S = \{1, 2, 3, 4, 6, 7\}$.

• Total digits available ($n$): 6
• Digits to be selected ($r$): 3
• Condition 1: The number must be even.
• Condition 2: No repetition of digits allowed.

[Per the Fundamental Principle of Counting, if an event can occur in $m$ ways and a second independent event can occur in $n$ ways, the total events occur in $m \times n$ ways.]

Step 2: Analyzing the Units Place (Constraint Application)

For a number to be even, the digit in the units place must be an even number. From the set $S = \{1, 2, 3, 4, 6, 7\}$, the available even digits are $\{2, 4, 6\}$.

Therefore, the units place can be filled in $3$ ways.

Step 3: Filling the Remaining Positions

Having fixed the units place, we select digits for the hundreds and tens places from the remaining digits.

• Total digits originally = 6.
• Digits used in units place = 1.
• Remaining digits = $6 - 1 = 5$.

Hundreds Place: Can be filled from the remaining 5 digits in $5$ ways. [By the Multiplication Principle of Counting].

Tens Place: Can be filled from the remaining 4 digits (as one was used for hundreds and one for units) in $4$ ways.

Step 4: Calculating Total Permutations

The total number of 3-digit even numbers is the product of the possibilities for each position:

$\text{Total} = (\text{Choices for Hundreds}) \times (\text{Choices for Tens}) \times (\text{Choices for Units})$

$\text{Total} = 5 \times 4 \times 3$

$\text{Total} = 60$

Step 5: Tabular Verification of Selection Logic

Final Solution: The total number of 3-digit even numbers that can be formed is 60.