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:

Case	Girls (from 4)	Boys (from 9)	Calculation
Case 1: 0 Girls	$\binom{4}{0}$	$\binom{9}{7}$	$\binom{4}{0} \times \binom{9}{7}$
Case 2: 1 Girl	$\binom{4}{1}$	$\binom{9}{6}$	$\binom{4}{1} \times \binom{9}{6}$
Case 3: 2 Girls	$\binom{4}{2}$	$\binom{9}{5}$	$\binom{4}{2} \times \binom{9}{5}$
Case 4: 3 Girls	$\binom{4}{3}$	$\binom{9}{4}$	$\binom{4}{3} \times \binom{9}{4}$

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: 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: 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.

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 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).

[Per the Fundamental Principle of Counting]

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

$^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10$

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

$^{21}C_2 = \frac{21!}{2!(21-2)!} = \frac{21 \times 20}{2 \times 1} = 210$

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 $n!$:

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

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:

$\text{Total Words} = (\text{Ways to Select}) \times (\text{Ways to Arrange})$ $\text{Total Words} = 2100 \times 24$ $\text{Total Words} = 50400$

Final Solution: 50,400 words

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:

Case	Questions from Part I ($r_1$)	Questions from Part II ($r_2$)	Total ($r_1 + r_2$)
Case 1	$3$	$5$	$8$
Case 2	$4$	$4$	$8$
Case 3	$5$	$3$	$8$

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 $P_2$):

$_5C_3 \times _7C_5 = \frac{5!}{3!2!} \times \frac{7!}{5!2!} = 10 \times 21 = 210$

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

$_5C_4 \times _7C_4 = \frac{5!}{4!1!} \times \frac{7!}{4!3!} = 5 \times 35 = 175$

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

$_5C_5 \times _7C_3 = 1 \times \frac{7!}{3!4!} = 1 \times 35 = 35$

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.

Step 1: Identifying the Frequency of Letters in 'EXAMINATION'

The word 'EXAMINATION' consists of 11 letters. We identify the frequency of each distinct letter as follows:

• Total letters ($n$) = 11
• E: 1
• X: 1
• A: 2
• M: 1
• I: 2
• N: 2
• T: 1
• O: 1

[By the Fundamental Principle of Counting and Permutations of Multisets, the total number of arrangements is given by $\frac{n!}{n_1! n_2! \dots n_k!}$].

Step 2: Determining the Alphabetical Order

To list the words in dictionary order, we arrange the unique letters of 'EXAMINATION' alphabetically:

A, E, I, M, N, O, T, X

Step 3: Calculating Words Before 'E'

A word in the dictionary starts with 'E' only after all words starting with 'A' have been exhausted. Therefore, we must calculate the total number of distinct arrangements that begin with the letter 'A'.

When the first position is fixed as 'A', we have 10 remaining positions to fill with the letters: {E, X, A, M, I, N, A, T, I, O, N}. Note: One 'A' is used, leaving us with:

• Remaining letters (10): E(1), X(1), A(1), M(1), I(2), N(2), T(1), O(1)

Step 4: Applying the Permutation Formula for Multiset

The number of ways to arrange the remaining 10 letters is given by the formula:

$$ \text{Arrangements} = \frac{10!}{n_E! \cdot n_X! \cdot n_A! \cdot n_M! \cdot n_I! \cdot n_N! \cdot n_T! \cdot n_O!} $$

Substituting the frequencies: $n_E=1, n_X=1, n_A=1, n_M=1, n_I=2, n_N=2, n_T=1, n_O=1$

$$ \text{Arrangements} = \frac{10!}{1! \cdot 1! \cdot 1! \cdot 1! \cdot 2! \cdot 2! \cdot 1! \cdot 1!} $$

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

$$ \text{Arrangements} = \frac{3,628,800}{4} = 907,200 $$

Step 5: Visualizing the Dictionary Placement

The number of words starting with 'A' represent all dictionary entries that appear strictly before the first entry starting with 'E'.

Final Solution: 907,200

Step 1: Identifying the Constraints

We are tasked with forming a 6-digit number using the set of digits $S = \{0, 1, 3, 5, 7, 9\}$ such that:

• The number must be divisible by $10$. [A number is divisible by $10$ if and only if its units digit is $0$].
• No digit is repeated. [Since there are exactly 6 unique digits provided and we must form a 6-digit number, we must use each digit exactly once].

Step 2: Defining the Positional Requirements

A 6-digit number has six positions: $P_6 P_5 P_4 P_3 P_2 P_1$.

Position	Constraint
$P_1$ (Units Place)	Must be $0$ for divisibility by $10$.
$P_6, P_5, P_4, P_3, P_2$	Remaining digits $\{1, 3, 5, 7, 9\}$ must occupy these positions.

Step 3: Calculating Permutations for Each Position

• Fixing the Units Place ($P_1$): There is only $1$ way to place the digit $0$. [Constraint: Divisibility by $10$ rule].
• Filling the Remaining Positions ($P_6$ to $P_2$): We have $5$ remaining digits $\{1, 3, 5, 7, 9\}$ to be arranged in the remaining $5$ positions. The number of ways to arrange $n$ distinct objects in $n$ positions is given by $n!$ [By Fundamental Counting Principle].

Step 4: Algebraic Derivation

Total numbers $N$ is calculated as:

$N = (\text{Choices for } P_1) \times (\text{Permutations of remaining } 5 \text{ digits in } 5 \text{ positions})$

$N = 1 \times 5!$

$N = 1 \times (5 \times 4 \times 3 \times 2 \times 1)$

$N = 120$

Step 5: Verification of Digit Constraints

Since the set $S$ contains $0$, we must ensure $0$ is not in the $P_6$ (leading) position. However, since $P_1$ is fixed as $0$, the digit $0$ cannot occupy $P_6$. Thus, all $120$ permutations satisfy the criteria for a valid 6-digit number.

Final Solution: 120

Step 1: Analyzing the Selection Constraints

The total population is $9$ boys and $4$ girls. We are required to form a committee of $7$ members. The constraint is that the committee must contain at least 3 girls.

Given the constraint of having at least $3$ girls in a $7$-member committee, we identify the valid cases based on the number of girls ($G$) and boys ($B$):

• Case 1: $3$ girls and $4$ boys.
• Case 2: $4$ girls and $3$ boys.

[By the Fundamental Counting Principle and the definition of Combinations, where selecting $r$ items from $n$ is given by $^nC_r = \frac{n!}{r!(n-r)!}$]

Step 2: Calculating Combinations for Each Case

Case 1: Selecting 3 girls and 4 boys

Number of ways to select $3$ girls from $4$ available girls: $^4C_3 = \frac{4!}{3!(4-3)!} = 4$.

Number of ways to select $4$ boys from $9$ available boys: $^9C_4 = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 126$.

Total for Case 1: $4 \times 126 = 504$.

Case 2: Selecting 4 girls and 3 boys

Number of ways to select $4$ girls from $4$ available girls: $^4C_4 = \frac{4!}{4!(4-4)!} = 1$.

Number of ways to select $3$ boys from $9$ available boys: $^9C_3 = \frac{9!}{3!(9-3)!} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84$.

Total for Case 2: $1 \times 84 = 84$.

Step 3: Summation of Mutually Exclusive Cases

Since these cases are mutually exclusive, we apply the Addition Principle of counting:

Total Ways = (Ways for Case 1) + (Ways for Case 2)

Total Ways = $504 + 84 = 588$.

Final Solution: The total number of ways to form the committee is 588.

Step 1: Identifying the Mathematical Framework

To determine the number of ways to form a committee of 7 members from a pool of 13 individuals (9 boys and 4 girls), where order of selection does not matter, we utilize the concept of Combinations. The number of ways to choose $r$ objects from a set of $n$ distinct objects is given by the formula [Per the Fundamental Principle of Combinations]:

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

Step 2: Defining Constraints and Variables

The total committee size is fixed at $7$. The constraint for this specific case is that the committee must contain exactly 3 girls.

• Total Girls Available ($n_g$) = $4$
• Number of Girls Required ($r_g$) = $3$
• Total Boys Available ($n_b$) = $9$
• Number of Boys Required ($r_b$) = Total Committee Size - Girls Required = $7 - 3 = 4$

Step 3: Calculating Independent Selections

We calculate the number of ways to select the girls and the boys separately:

Part A: Selecting 3 girls from 4

$$\binom{4}{3} = \frac{4!}{3!(4-3)!} = \frac{4 \times 3!}{3! \times 1!} = 4$$

Part B: Selecting 4 boys from 9

$$\binom{9}{4} = \frac{9!}{4!(9-4)!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{ (4 \times 3 \times 2 \times 1) \times 5!} = \frac{3024}{24} = 126$$

Step 4: Applying the Fundamental Counting Principle

Since the selection of girls and the selection of boys are independent events occurring simultaneously to complete the committee, we apply the Multiplication Principle [Per the Fundamental Counting Principle]:

$$\text{Total Ways} = \binom{4}{3} \times \binom{9}{4}$$

$$\text{Total Ways} = 4 \times 126 = 504$$

Visual Representation of Combinatorial Logic

Final Solution: The committee can be formed in 504 ways.

Step 1: Decomposition of the Given Word

The word EQUATION consists of $8$ distinct letters:

• Vowels ($V$): {E, U, A, I, O} — Total $n_v = 5$
• Consonants ($C$): {Q, T, N} — Total $n_c = 3$

[Per the Fundamental Principle of Counting, since the letters are distinct, we treat them as individual entities for arrangement.]

Step 2: Grouping Strategy

The problem requires vowels to occur together and consonants to occur together. We treat the set of vowels as one single unit ($V_{unit}$) and the set of consonants as one single unit ($C_{unit}$).

Total items to arrange = $\{V_{unit}, C_{unit}\}$.

Number of ways to arrange these 2 units = $2!$ [By the Permutation Theorem $n!$ for $n$ distinct items].

Step 3: Internal Permutations

A. Arranging Vowels internally: The 5 distinct vowels {E, U, A, I, O} can be arranged among themselves in $5!$ ways.

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

B. Arranging Consonants internally: The 3 distinct consonants {Q, T, N} can be arranged among themselves in $3!$ ways.

Calculation: $3! = 3 \times 2 \times 1 = 6$ ways.

Step 4: Total Calculation

Applying the Multiplication Principle [If an event can occur in $m$ ways and another independent event in $n$ ways, the total number of occurrences is $m \times n$]:

Total Words = (Ways to arrange units) $\times$ (Internal arrangement of Vowels) $\times$ (Internal arrangement of Consonants)

Total Words = $2! \times 5! \times 3!$

Total Words = $2 \times 120 \times 6$

Total Words = $1440$

Final Solution: The total number of words that can be formed is $1440$.

Step 1: Analyzing the Selection Criteria

The problem requires the selection of a 5-card combination from a standard deck of 52 cards, subject to the constraint that exactly one card must be a King. A standard deck contains:

• Total cards: $52$
• Total Kings: $4$
• Total Non-Kings: $48$

[Per the Fundamental Principle of Counting and the definition of Combinations, the total selection is the product of independent selections from mutually exclusive subsets.]

Step 2: Formulating the Combination Components

To satisfy the requirement of exactly one King, the selection must be split into two distinct parts:

1. Selection of the King: We must choose $1$ King from the $4$ available Kings. This is denoted as ${^4C_1}$.
2. Selection of the remaining cards: Since a 5-card hand is required and $1$ card is already a King, $4$ additional cards must be chosen from the remaining non-king cards ($48$). This is denoted as ${^{48}C_4}$.

Step 3: Algebraic Calculation

The total number of combinations is given by the product: ${^4C_1} \times {^{48}C_4}$.

Using the combination formula ${^nC_r} = \frac{n!}{r!(n-r)!}$:

${^4C_1} = \frac{4!}{1!(4-1)!} = \frac{4}{1} = 4$

${^{48}C_4} = \frac{48!}{4!(48-4)!} = \frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1}$

Step 4: Simplifying the Calculation

${^{48}C_4} = \frac{48}{24} \times 47 \times 46 \times 45 = 2 \times 47 \times 46 \times 45$

${^{48}C_4} = 194,580$

Now, multiply by the selection of the King:

Total Combinations $= 4 \times 194,580 = 778,320$

Visual Representation of Selection Logic

Final Solution: 778,320