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

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

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.

Step 1: Analyzing the Selection Criteria

The problem requires the selection of a total of $9$ balls from three distinct groups of colored balls. The constraints are defined as follows:

• Total red balls available ($n_R$) = $6$
• Total white balls available ($n_W$) = $5$
• Total blue balls available ($n_B$) = $5$
• Selection requirement: Exactly $3$ balls of each color.

Step 2: Mathematical Formulation using Combinations

Since the order of selection does not matter within the color groups, we employ the combination formula, denoted as $^nC_r = \frac{n!}{r!(n-r)!}$, where $n$ is the total number of items to choose from and $r$ is the number of items chosen [Per the Principle of Combinatorics].

Step 3: Calculating Individual Selections

We calculate the number of ways to select 3 balls for each color independently:

(i) Selection of 3 red balls from 6:

$$^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$$

(ii) Selection of 3 white balls from 5:

$$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$

(iii) Selection of 3 blue balls from 5:

$$^5C_3 = \frac{5!}{3!(5-3)!} = \frac{5 \times 4}{2 \times 1} = 10$$

Step 4: Applying the Fundamental Counting Principle

According to the Fundamental Counting Principle (Multiplication Principle), if one task can be done in $m$ ways and a second task can be done in $n$ ways, then both tasks can be performed in $m \times n$ ways. Since the selections of red, white, and blue balls are independent operations, we multiply the individual results [Per the Fundamental Principle of Counting].

Total Ways = $^6C_3 \times ^5C_3 \times ^5C_3$

Total Ways = $20 \times 10 \times 10$

Final Solution: 2000 ways

Step 1: Definition of the Combinatorial Formula

To solve for $n$, we utilize the standard formula for combinations, which defines the number of ways to select $r$ objects from a set of $n$ objects [Per the definition of $nCr$]:

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

Step 2: Expanding the Given Ratio

Given the equation ${}^{2n}C_3 : {}^nC_3 = 11 : 1$, we write this as a fractional expression:

$\frac{{}^{2n}C_3}{{}^nC_3} = \frac{11}{1}$

Applying the factorial formula to both terms:

$\frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11$

Step 3: Algebraic Simplification

Cancel the constant $3!$ from both numerator and denominator [By the properties of algebraic fractions]:

$\frac{(2n)!}{(2n-3)!} \times \frac{(n-3)!}{n!} = 11$

Expand the factorials to eliminate $(2n-3)!$ and $(n-3)!$:

$\frac{(2n)(2n-1)(2n-2)(2n-3)!}{(2n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!} = 11$

$\frac{(2n)(2n-1)(2n-2)}{n(n-1)(n-2)} = 11$

Step 4: Solving for $n$

Factor out common terms from the numerator:

$\frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11$

Cancel $n$ and $(n-1)$ from the numerator and denominator [Condition: $n > 3$ as per the definition of $nC_r$]:

$\frac{4(2n-1)}{n-2} = 11$

Multiply to solve the linear equation:

$4(2n - 1) = 11(n - 2)$

$8n - 4 = 11n - 22$

$22 - 4 = 11n - 8n$

$18 = 3n$

$n = 6$

Step 5: Verification of Geometric/Combinatorial Constraints

Final Solution: $n = 6$

Step 1: Formulaic Representation and Given Constraints

The definition of the combination formula is given by ${}^{n}C_r = \frac{n!}{r!(n-r)!}$, where $n \geq r \geq 0$. We are provided with the ratio:

$\frac{{}^{2n}C_3}{{}^{n}C_3} = \frac{12}{1}$

Step 2: Expansion of the Combination Terms

Applying the formula to both components of the ratio:

${}^{2n}C_3 = \frac{(2n)!}{3!(2n-3)!} = \frac{(2n)(2n-1)(2n-2)(2n-3)!}{6(2n-3)!} = \frac{2n(2n-1)(2n-2)}{6}$

${}^{n}C_3 = \frac{n!}{3!(n-3)!} = \frac{n(n-1)(n-2)(n-3)!}{6(n-3)!} = \frac{n(n-1)(n-2)}{6}$

Step 3: Algebraic Setup and Simplification

Substituting these expressions back into the ratio equation:

$\frac{\left[ \frac{2n(2n-1)(2n-2)}{6} \right]}{\left[ \frac{n(n-1)(n-2)}{6} \right]} = 12$

[Per the property of division of fractions, the denominators (6) cancel out]:

$\frac{2n(2n-1)(2n-2)}{n(n-1)(n-2)} = 12$

Step 4: Factoring and Reduction

Observe that $(2n-2) = 2(n-1)$. Substitute this into the numerator:

$\frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 12$

[By cancelling common factors $n$ and $(n-1)$ from the numerator and denominator, noting $n > 2$]:

$\frac{4(2n-1)}{(n-2)} = 12$

Step 5: Solving for $n$

Divide both sides by 4:

$\frac{2n-1}{n-2} = 3$

[By cross-multiplication]:

$2n - 1 = 3(n - 2)$

$2n - 1 = 3n - 6$

[Rearranging terms to solve for $n$]:

$6 - 1 = 3n - 2n$

$n = 5$

Verification Diagram

Final Solution: n = 5

Step 1: Theoretical Foundation & Formulation

To determine the number of ways to select a specific committee or team where the order of selection is irrelevant, we employ the Combination formula. The number of ways to select $r$ items from a set of $n$ distinct items is defined as:

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

[Per the Fundamental Principle of Combinatorics for selection without arrangement]

Step 2: Decomposition of Selection Tasks

The problem requires two independent selection processes to be performed simultaneously to form the team:

• Task A: Selecting 3 boys from a group of 5 boys.
• Task B: Selecting 3 girls from a group of 4 girls.

Step 3: Algebraic Calculation of Selections

Applying the formula to Task A ($n=5, r=3$):

$^{5}C_{3} = \frac{5!}{3!(5-3)!} = \frac{5 \times 4 \times 3!}{3! \times 2 \times 1} = \frac{20}{2} = 10$

Applying the formula to Task B ($n=4, r=3$):

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

Step 4: Applying the Fundamental Counting Principle

According to the Fundamental Principle of Multiplication, if one operation can be performed in $m$ ways and a second independent operation in $n$ ways, the total number of ways to perform both operations is $m \times n$.

Total ways = ($^{5}C_{3}$) $\times$ ($^{4}C_{3}$)

Total ways = $10 \times 4 = 40$

Visual Representation of Selection Logic

Final Solution: The total number of ways to select the team is 40 ways.

Step 1: Application of the Complementary Combination Theorem

The problem presents the equation ${}^nC_8 = {}^nC_2$. We utilize the fundamental theorem of combinations which states that if ${}^nC_a = {}^nC_b$, then either $a = b$ or $n = a + b$.

[Per the property: ${}^nC_r = {}^nC_{n-r}$]

Since $8 \neq 2$, we must satisfy the condition:

$n = 8 + 2$

$n = 10$

Step 2: Verification of Variable $n$

With $n = 10$, we check the equality:

${}^{10}C_8 = \frac{10!}{8!(10-8)!} = \frac{10 \times 9}{2 \times 1} = 45$

${}^{10}C_2 = \frac{10!}{2!(10-2)!} = \frac{10 \times 9}{2 \times 1} = 45$

[Verified: ${}^{10}C_8 = {}^{10}C_2$ holds true]

Step 3: Calculating the Required Value ${}^nC_2$

Substituting $n = 10$ into the expression ${}^nC_2$:

${}^{10}C_2 = \frac{n!}{r!(n-r)!} = \frac{10!}{2!(10-2)!}$

${}^{10}C_2 = \frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}$

${}^{10}C_2 = \frac{90}{2} = 45$

Visual Representation of Combinatorial Distribution

Final Solution: ${}^{n}C_2 = 45$

Step 1: Analyzing the Selection Parameters

To determine the number of ways to select balls of specific colors, we employ the principle of combinations, as the order of selection does not matter [Per the definition of combinations: $^nC_r = \frac{n!}{r!(n-r)!}$].

Given parameters:

• Total number of black balls ($n_b$) = $5$
• Number of black balls to be selected ($r_b$) = $2$
• Total number of red balls ($n_r$) = $6$
• Number of red balls to be selected ($r_r$) = $3$

Step 2: Calculating Combinations for Each Color

First, calculate the number of ways to select 2 black balls from 5 available [By the Combination Formula]:

$$^5C_2 = \frac{5!}{2!(5-2)!} = \frac{5 \times 4 \times 3!}{2 \times 1 \times 3!} = \frac{20}{2} = 10$$

Second, calculate the number of ways to select 3 red balls from 6 available:

$$^6C_3 = \frac{6!}{3!(6-3)!} = \frac{6 \times 5 \times 4 \times 3!}{ (3 \times 2 \times 1) \times 3!} = \frac{120}{6} = 20$$

Step 3: Applying the Fundamental Principle of Counting

Since the selection of black balls and the selection of red balls are independent events that must both occur to satisfy the condition, we use the Fundamental Principle of Multiplication [Per the Fundamental Counting Principle (Multiplication Rule)]:

$$\text{Total Ways} = (^5C_2) \times (^6C_3)$$

$$\text{Total Ways} = 10 \times 20 = 200$$

Final Solution: The total number of ways to select the balls is 200.

Step 1: Identifying the Mathematical Framework

To determine the number of ways to form the team, we apply the Fundamental Counting Principle and the Combinations formula. Since the order of selection within the team does not matter, we use the formula for combinations denoted as $^nC_r = \frac{n!}{r!(n-r)!}$. [Per the Principle of Combinations]

Step 2: Categorization of Players

The total pool of 17 players is partitioned into two distinct groups based on the capability to bowl:

• Group A (Bowlers): $n_1 = 5$ players.
• Group B (Non-bowlers): $n_2 = 17 - 5 = 12$ players.

Step 3: Defining Selection Constraints

A team of 11 players must be formed with exactly 4 bowlers. This implies:

• Selection from Group A (Bowlers): Select $r_1 = 4$ out of $n_1 = 5$.
• Selection from Group B (Non-bowlers): Since the total team size is 11, the remaining spots $r_2 = 11 - 4 = 7$ must be filled from the $n_2 = 12$ non-bowlers.

Step 4: Calculating Individual Combinations

Selection of Bowlers ($^5C_4$):

$^5C_4 = \frac{5!}{4!(5-4)!} = \frac{5 \times 4!}{4! \times 1!} = 5$

Selection of Non-bowlers ($^{12}C_7$):

$^{12}C_7 = \frac{12!}{7!(12-7)!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5!}$

$^{12}C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95040}{120} = 792$

Step 5: Applying the Multiplication Principle

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

Total ways = $^5C_4 \times ^{12}C_7$

Total ways = $5 \times 792 = 3960$

Visual Representation of Combinatorial Selection

Final Solution: The total number of ways to select the team is 3960.