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

Given Variables & Initial Setup

Total pool of candidates: 9 boys ($n_B = 9$) and 4 girls ($n_G = 4$).

Total committee size to be formed: $r = 7$.

Constraint: The committee must contain at most 3 girls. This means the number of girls ($g$) can be 0, 1, 2, or 3.

Step 1: Establishing the Combinatorial Formula

To choose $r$ items from $n$ distinct items, we use the combination formula: $C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}$ [Per the Fundamental Principle of Counting].

Step 2: Defining Exhaustive Mutually Exclusive Cases

Since we require "at most 3 girls" in a committee of 7, we partition the problem into the following disjoint cases:

Step 3: Performing Calculations

• Case 1: $\binom{4}{0} \times \binom{9}{7} = 1 \times \binom{9}{2} = 1 \times \frac{9 \times 8}{2 \times 1} = 36$
• Case 2: $\binom{4}{1} \times \binom{9}{6} = 4 \times \binom{9}{3} = 4 \times \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 4 \times 84 = 336$
• Case 3: $\binom{4}{2} \times \binom{9}{5} = \binom{4}{2} \times \binom{9}{4} = 6 \times \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 6 \times 126 = 756$
• Case 4: $\binom{4}{3} \times \binom{9}{4} = 4 \times 126 = 504$

Step 4: Summation of Mutually Exclusive Cases

Total ways = $36 + 336 + 756 + 504$ [Per the Addition Principle of Combinatorics].

Total ways = $1632$.

Visualization of Selection Constraints

Final Solution: 1632 ways

Step 1: Analyzing the Word Composition

The given word is ASSASSINATION. First, we identify the frequency of each distinct letter present in the word:

• A: 3
• S: 4
• I: 2
• N: 2
• T: 1
• O: 1

The total number of letters $n = 13$. [Based on the Fundamental Counting Principle for multiset permutations].

Step 2: Grouping the Constraints

The condition requires all $S$'s to be together. We treat the block of four $S$'s $(SSSS)$ as a single super-letter or entity, denoted as $X$.

The remaining letters are: $A, A, A, I, I, N, N, T, O$.

Total entities to arrange = $\{X, A, A, A, I, I, N, N, T, O\}$.

Number of entities = $1 + 3 + 2 + 2 + 1 + 1 = 10$.

Step 3: Calculating Permutations of the Entities

We must arrange these 10 entities, accounting for the repetitions of $A$ (3 times), $I$ (2 times), and $N$ (2 times). [Per the formula for permutations of a multiset: $\frac{n!}{n_1! n_2! ... n_k!}$].

The number of arrangements is given by:

$$P = \frac{10!}{3! \times 2! \times 2!}$$

Calculating the values:

$10! = 3,628,800$

$3! = 6$

$2! = 2$

$P = \frac{3,628,800}{6 \times 2 \times 2} = \frac{3,628,800}{24}$

$P = 151,200$

Step 4: Internal Arrangement of the S-Block

Since all four $S$'s are identical, the number of ways to arrange them within the block $X$ is $\frac{4!}{4!} = 1$. [By the logic of permutations of identical objects].

Step 5: Final Computation

The total number of arrangements is the product of the arrangement of the entities and the internal arrangements of the block.

$$\text{Total Ways} = 151,200 \times 1 = 151,200$$

Final Solution: 151,200 ways

Step 1: Theoretical Foundation

A chord of a circle is defined as a straight line segment whose endpoints both lie on the circumference of the circle. To form a unique chord, exactly two distinct points on the circle must be selected [Per the geometric definition of a chord]. Given a set of $n$ distinct points on a circle, the number of ways to choose $2$ points from the set of $n$ points is given by the combination formula:

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

Where:

• $n = 21$ (Total number of points available)
• $r = 2$ (Number of points required to define a chord)

Step 2: Mathematical Derivation

Substituting the given variables into the combination formula:

$$\text{Number of chords} = ^{21}C_2 = \frac{21!}{2!(21-2)!}$$

Expanding the factorials for simplification:

$$\text{Number of chords} = \frac{21 \times 20 \times 19!}{2 \times 1 \times 19!}$$

Canceling the common $19!$ terms in the numerator and denominator [By the multiplicative property of factorials]:

$$\text{Number of chords} = \frac{21 \times 20}{2}$$

$$\text{Number of chords} = 21 \times 10 = 210$$

Step 3: Geometric Representation

Step 4: Analytical Conclusion

Since the points are on a circle, no three points are collinear. Therefore, every unique pair of points selected from the set of 21 will result in a unique chord. No geometric constraints (such as collinearity) reduce the total count calculated via the combination formula.

Final Solution: 210 chords

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 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: Definition of Factorial Notation

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. The definition is given as: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$ [By the fundamental definition of factorials in combinatorics].

Step 2: Calculation of Left-Hand Side (LHS)

The expression on the left-hand side is $3! + 4!$.

• Calculate $3!$: $3! = 3 \times 2 \times 1 = 6$
• Calculate $4!$: $4! = 4 \times 3 \times 2 \times 1 = 24$

Summing these values: $3! + 4! = 6 + 24 = 30$ [Arithmetic addition].

Step 3: Calculation of Right-Hand Side (RHS)

The expression on the right-hand side is $7!$.

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

Performing the multiplication sequentially: $7 \times 6 = 42$ $42 \times 5 = 210$ $210 \times 4 = 840$ $840 \times 3 = 2520$ $2520 \times 2 = 5040$ $5040 \times 1 = 5040$ [Arithmetic multiplication].

Step 4: Comparison and Analysis

Comparing the calculated values: $30 \neq 5040$.

Step 5: Theoretical Generalization

In general, for any positive integers $m$ and $n$ where $m, n > 1$, $(m! + n!) \neq (m+n)!$ because the factorial function grows super-exponentially. Specifically, by the properties of factorials: $(m+n)! = (m+n) \times (m+n-1) \times \dots \times (n+1) \times n!$ This product is significantly larger than the sum of two individual factorials.

Final Solution: No, $3! + 4! \neq 7!$ because $30 \neq 5040$.

Step 1: Analyzing the Source Word

The word "DAUGHTER" consists of the following 8 unique letters: {D, A, U, G, H, T, E, R}.

• Vowels in the set: {A, U, E} (Total count = 3)
• Consonants in the set: {D, G, H, T, R} (Total count = 5)

Step 2: Selection Process (Combinations)

To form a word with 2 vowels and 3 consonants, we must perform two independent selection tasks [Per the Fundamental Counting Principle of Selection]:

• Selecting 2 vowels from 3: $^3C_2 = \frac{3!}{2!(3-2)!} = 3$ ways.
• Selecting 3 consonants from 5: $^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$ ways.

Total combinations of letters = $3 \times 10 = 30$ ways.

Step 3: Arrangement Process (Permutations)

Each set of 5 selected letters can be arranged among themselves to form different words. The number of permutations of $n$ distinct objects is given by $n!$ [Per the Permutation Theorem].

For each selection of 5 letters, the number of ways to arrange them is $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ ways.

Step 4: Total Permutations Calculation

Using the Multiplication Principle for counting, the total number of words that can be formed is the product of the number of ways to select the letters and the number of ways to arrange those selected letters:

Total Words = $(\text{Ways to Select}) \times (\text{Ways to Arrange})$

Total Words = $30 \times 120 = 3,600$.

Final Solution: 3,600 words

Step 1: Theoretical Foundation of Factorials

The factorial of a non-negative integer $n$, denoted by $n!$, is the product of all positive integers less than or equal to $n$. It is defined by the recurrence relation: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$. [Per the Fundamental Definition of Factorials].

Step 2: Decomposition of the Factorial Expression

To evaluate the expression $\frac{8!}{6! \times 2!}$, we utilize the property of factorials where $n! = n \times (n-1) \times (n-2)!$. This allows for the cancellation of common terms in the numerator and denominator.

Expanding the numerator until we reach the largest factorial in the denominator:

$8! = 8 \times 7 \times 6!$

Step 3: Algebraic Substitution and Simplification

Substitute the expanded numerator back into the original expression:

$\text{Expression} = \frac{8 \times 7 \times 6!}{6! \times 2!}$

Cancel the common term $6!$ from the numerator and the denominator: [By the Property of Multiplicative Identity and Division of Like Terms]

$\text{Expression} = \frac{8 \times 7}{2!}$

Step 4: Final Calculation

Evaluate the denominator and perform the final multiplication:

• $2! = 2 \times 1 = 2$
• $\text{Expression} = \frac{56}{2}$
• $\text{Expression} = 28$

Visual Representation of Factorial Reduction

Final Solution: 28

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: Expanding the Factorial Terms

Given the equation: $$\frac{1}{6!} + \frac{1}{7!} = \frac{x}{8!}$$ To simplify the expression, we utilize the property of factorials where $n! = n \times (n-1)!$ [Per the definition of factorial functions]. We express the larger factorials in terms of $6!$:

• $7! = 7 \times 6!$
• $8! = 8 \times 7 \times 6! = 56 \times 6!$

Step 2: Substituting and Simplifying the Denominators

Substitute the expanded factorials into the equation:

To eliminate the common factor of $\frac{1}{6!}$ from the entire equation, we multiply every term by $6!$ [By the Multiplicative Property of Equality]:

Step 3: Algebraic Evaluation of the Left-Hand Side (LHS)

Evaluate the addition on the LHS by finding a common denominator:

Step 4: Isolating the Variable $x$

Multiply both sides of the equation by $56$ to solve for $x$ [By the Division/Multiplication Property of Equality]:

Visual Representation of the Factorial Relationship

Final Solution: The value of $x$ is $64$.

Step 1: Analyzing the Positional Constraints

We are required to arrange a total of $n = 9$ individuals (5 men and 4 women) in a single row. The constraint specifies that the 4 women must occupy the even-numbered positions in the row [Given condition].

The available positions in the row are: $P_1, P_2, P_3, P_4, P_5, P_6, P_7, P_8, P_9$.

The even positions within this set of nine are: $P_2, P_4, P_6, P_8$. There are exactly 4 even places available, which matches the number of women to be seated [Logical parity match].

Step 2: Calculating Arrangements for Women

There are 4 women and 4 designated even positions. The number of ways to arrange $n$ distinct objects in $n$ positions is given by the factorial $n!$ [Fundamental Principle of Counting].

Number of ways to seat the women $= 4!$

$4! = 4 \times 3 \times 2 \times 1 = 24$ ways.

Step 3: Calculating Arrangements for Men

After the women are seated in the even positions, there remain $9 - 4 = 5$ odd positions for the 5 men ($P_1, P_3, P_5, P_7, P_9$).

The number of ways to arrange 5 distinct men in 5 distinct positions is given by $5!$ [Permutation principle].

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

Step 4: Applying the Fundamental Principle of Multiplication

According to the Fundamental Counting Principle, if one event can occur in $m$ ways and a second independent event can occur in $n$ ways, then the total number of ways the two events can occur in sequence is $m \times n$ [Multiplication Axiom].

Total arrangements $= (\text{Ways to arrange women}) \times (\text{Ways to arrange men})$

Total arrangements $= 4! \times 5!$

Total arrangements $= 24 \times 120 = 2880$.

Visual Representation of Occupancy

Final Solution: The total number of possible arrangements is 2880.

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

Given Variables & Initial Setup

• Total number of courses available ($n_{total}$) = $9$
• Number of courses to be chosen ($r_{total}$) = $5$
• Number of specific compulsory courses ($n_{comp}$) = $2$

Since the 2 specific courses are compulsory, they must be included in every student's selection. This effectively reduces the number of courses the student needs to "choose" and the number of "options" available for selection.

Step 1: Calculating Remaining Selections

Because $2$ courses are already fixed, the number of courses remaining to be chosen is:

$r_{rem} = r_{total} - n_{comp} = 5 - 2 = 3$

[By the fundamental principle of subtraction in selection processes]

Step 2: Calculating Remaining Options

The number of available courses from which the student can make the remaining selection is:

$n_{rem} = n_{total} - n_{comp} = 9 - 2 = 7$

Step 3: Combinatorial Calculation

The problem now reduces to selecting $3$ courses out of the remaining $7$ available courses. Since the order of selection does not matter for a programme of courses, we use the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$ [Per the definition of combinations].

Substituting the values:

$^7C_3 = \frac{7!}{3!(7-3)!}$

$^7C_3 = \frac{7!}{3! \times 4!}$

$^7C_3 = \frac{7 \times 6 \times 5 \times 4!}{ (3 \times 2 \times 1) \times 4!}$

$^7C_3 = \frac{7 \times 6 \times 5}{6}$

$^7C_3 = 7 \times 5 = 35$

Visual Representation of Selection Logic

Comparison Table of Selection Logic

Final Solution: 35 ways

Given Variables & Initial Setup

Total number of students ($n$) = $25$.

Number of students to be chosen ($r$) = $10$.

Special group size ($k$) = $3$. These students have a condition: either all 3 join or none join.

This problem is partitioned into two mutually exclusive cases [Per the Addition Principle of Counting].

Step 1: Case I - All 3 students join the excursion

If all 3 specific students join the party, we have already filled $3$ spots out of the $10$ required spots.

Remaining spots to be filled = $10 - 3 = 7$.

Remaining pool of students available = $25 - 3 = 22$.

The number of ways to choose $7$ students from $22$ is given by the combination formula $_nC_r = \frac{n!}{r!(n-r)!}$.

Calculation: $_{22}C_7 = \frac{22!}{7!(22-7)!} = \frac{22!}{7! \times 15!}$

Substituting the values: $\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 170,544$

Step 2: Case II - None of the 3 students join the excursion

If none of the 3 students join, we must fill all $10$ spots from the remaining pool of students.

Remaining pool of students = $25 - 3 = 22$.

Number of spots to be filled = $10$.

Calculation: $_{22}C_{10} = \frac{22!}{10!(22-10)!} = \frac{22!}{10! \times 12!}$

Substituting the values: $\frac{22 \times 21 \times 20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13}{10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 646,646$

Step 3: Calculating Total Combinations

According to the Fundamental Principle of Addition, since these two cases are mutually exclusive [Per the Addition Principle], we add the results of both cases:

$\text{Total ways} = (_{22}C_7) + (_{22}C_{10})$

$\text{Total ways} = 170,544 + 646,646$

$\text{Total ways} = 817,190$

Final Solution: The total number of ways the excursion party can be chosen is 817,190.

Step 1: Identifying the Mathematical Expression

The problem requires the evaluation of the factorial expression $\frac{n!}{(n-r)!}$ given the parameters $n = 6$ and $r = 2$. This expression represents the formula for calculating permutations of $n$ items taken $r$ at a time, denoted as $P(n, r) = \frac{n!}{(n-r)!}$. [By the Fundamental Principle of Counting]

Step 2: Substitution of Given Variables

Substitute $n = 6$ and $r = 2$ into the expression:

$\frac{6!}{(6-2)!}$

Simplifying the denominator: $6 - 2 = 4$.

The expression becomes: $\frac{6!}{4!}$

Step 3: Expansion of Factorials

Recall the definition of a factorial: $n! = n \times (n-1) \times (n-2) \times \dots \times 1$. [By the definition of the factorial function]

Expand $6!$ until reaching $4!$ to allow for cancellation:

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

$6! = 6 \times 5 \times 4!$

Step 4: Algebraic Simplification

Substitute the expanded form back into the quotient:

$\frac{6 \times 5 \times 4!}{4!}$

Cancel the common factor $4!$ from both the numerator and the denominator: [By the Property of Multiplicative Identity]

$6 \times 5 = 30$

Visual Representation of Permutation Calculation

Step 5: Tabular Summary of Computation

Final Solution: 30

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 Combinatorial Requirements

To form a word with $2$ distinct vowels and $2$ distinct consonants, we must execute the following logical sequence:

• Selection of vowels from the available pool.
• Selection of consonants from the available pool.
• Arrangement of the selected characters into a sequence (word).

Step 2: Selection of Characters

The total number of vowels is $n_v = 5$, and the number of consonants is $n_c = 21$. We are required to choose $r_v = 2$ vowels and $r_c = 2$ consonants.

The number of ways to select $2$ vowels from $5$ is given by the combination formula $^nC_r = \frac{n!}{r!(n-r)!}$:

The number of ways to select $2$ consonants from $21$ is:

Total ways to select the $4$ letters ($2$ vowels and $2$ consonants) is: $10 \times 210 = 2100$ [By the Multiplication Principle of Counting]

Step 3: Arrangement of Selected Characters

Once $4$ distinct letters are selected, they can be arranged among themselves to form a word of length $4$. The number of ways to arrange $n$ distinct objects is given by $n!$:

Step 4: Final Calculation

The total number of words that can be formed is the product of the number of ways to select the letters and the number of ways to arrange them:

Final Solution: 50,400 words

Step 1: Identifying the Mathematical Framework

To determine the number of combinations, we employ the combinatorial formula $^nC_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items available and $r$ is the number of items to be selected [Fundamental Counting Principle & Definition of Combinations].

Step 2: Partitioning the Set of Cards

A standard deck of 52 cards is partitioned into two distinct categories based on the problem constraints:

• Category A (Aces): There are $4$ aces in a standard deck.
• Category B (Non-Aces): The remaining cards in the deck are $52 - 4 = 48$ cards.

Step 3: Calculating Independent Selections

The requirement is a 5-card combination containing exactly one ace. This implies the selection process consists of two simultaneous events:

1. Selecting 1 Ace: We must choose 1 card from the 4 available aces.
   Calculation: $^4C_1 = \frac{4!}{1!(4-1)!} = 4$ [Property: $^nC_1 = n$]
2. Selecting the Remaining 4 Cards: Since the total hand must contain 5 cards and 1 is already an ace, we must choose 4 additional cards from the 48 non-ace cards.
   Calculation: $^{48}C_4 = \frac{48!}{4!(48-4)!} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$

Step 4: Executing Numerical Computation

Applying the division for the non-ace selection:

• $^{48}C_4 = \frac{4,669,920}{24} = 194,580$

By the Multiplication Principle of counting, the total number of combinations is the product of the two independent selections [Fundamental Principle of Counting]:

Total Combinations $= (^4C_1) \times (^{48}C_4)$

Total Combinations $= 4 \times 194,580$

Total Combinations $= 778,320$

Final Solution: The total number of 5-card combinations containing exactly one ace is 778,320.

Given Variables & Initial Setup

The total number of questions is $12$, divided as follows:

• Part I ($P_1$): $5$ questions available.
• Part II ($P_2$): $7$ questions available.

The student must select a total of $8$ questions such that the condition "at least 3 from each part" is satisfied. [Per the Fundamental Counting Principle and Selection Principle].

Step 1: Identifying Possible Selection Scenarios

Since the total number of questions $n = 8$ and the requirement is $\ge 3$ from $P_1$ and $\ge 3$ from $P_2$, we evaluate the combinations of $(r_1, r_2)$ where $r_1$ is the number of questions from Part I and $r_2$ is the number of questions from Part II:

Step 2: Calculating Combinations for Each Case

The number of ways to select $r$ items from $n$ is given by the combination formula $_nC_r = \frac{n!}{r!(n-r)!}$.

Calculation for Case 1 ($3$ from $P_1$, $5$ from $P_2$):

Calculation for Case 2 ($4$ from $P_1$, $4$ from $P_2$):

Calculation for Case 3 ($5$ from $P_1$, $3$ from $P_2$):

Step 3: Total Summation

To find the total number of ways, we apply the Addition Principle of counting [Per the Rule of Sum in Combinatorics]:

Total Ways = $210 + 175 + 35$

Total Ways = $420$

Visual Representation of Selection Logic

Final Solution: The student can select the questions in 420 ways.