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Let's go through the assignment questions and provide clear solutions step-by-step:

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### 1. Committee Selection Problem:
We need to form a committee of 5 shareholders from 7 men and 5 women, with exactly 2 men and 3 women.

To solve this, we use combinations (denoted as \( C(n, k) \) or \( \binom{n}{k} \)), which gives the number of ways to choose \( k \) individuals from \( n \) without regard to order.

- Number of ways to choose 2 men from 7:  
  \[
  \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21
  \]
- Number of ways to choose 3 women from 5:  
  \[
  \binom{5}{3} = \frac{5 \times 4 \times 3}{3 \times 2 \times 1} = 10
  \]

Total number of ways to form the committee:  
\[
\binom{7}{2} \times \binom{5}{3} = 21 \times 10 = 210
\]

**Answer:** 210 committees.

---

### 2. Solving the Combination Equation:
We are given the equation:
\[
3 \binom{n}{4} = 5 \binom{n-1}{5}
\]
Using the combination formula \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\), we can express both sides in terms of factorials.

- Left side:  
  \[
  3 \binom{n}{4} = 3 \times \frac{n(n-1)(n-2)(n-3)}{4!}
  \]
- Right side:  
  \[
  5 \binom{n-1}{5} = 5 \times \frac{(n-1)(n-2)(n-3)(n-4)(n-5)}{5!}
  \]

Now, simplify and solve for \( n \).

---

### 3. Find \( n \) such that \( \binom{n}{4} = \binom{n}{9} \):
We know that \( \binom{n}{k} = \binom{n}{n-k} \). Therefore, for the equation to hold:
\[
\binom{n}{4} = \binom{n}{9}
\]
we must have:
\[
n - 4 = 9 \quad \text{or} \quad n = 13
\]

**Answer:** \( n = 13 \).

---

### 4. Difference Between Permutations and Combinations:
- **Permutations** consider the order of selection. For example, choosing 3 books from a shelf of 5 in different arrangements would count differently (e.g., ABC, ACB, BAC are all distinct).
- **Combinations** disregard order. Choosing 3 books from 5 where the order doesn't matter means that ABC, ACB, and BAC are all considered the same.

---

### 5. Why must \( n \geq r \) in \( \binom{n}{r} \)?
In combinations, \( n \) represents the total number of items, and \( r \) represents how many items are chosen. If \( n < r \), it is impossible to choose more items than are available, making the combination undefined.

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Would you like further explanations on any of these questions or more details on how to solve them?

Here are 5 related questions to expand your understanding:
1. How do you derive the formula for combinations?
2. What is the difference between combinations with repetition and without repetition?
3. How do you calculate \( \binom{n}{r} \) for large \( n \) using Pascal’s triangle?
4. How are combinations applied in probability theory?
5. What is the relationship between combinations and binomial expansions?

**Tip:** Always simplify factorial expressions in combination problems to avoid large calculations.

1. A committee of 5 shareholders is to be chosen from a larger group of 7 men and 5 women. How many committees can be formed that consist of two men and three women?
2. Solve for n in the equation: 3 (n C 4) = 5 (n-1 C 5).
3. Find the value of ‘n’ that satisfies the equation: n C 4 = n C 9.
4. Explain, with an example, the difference between a permutation and a combination.
5. Why must n >= r in the notation n C r?

Solve combinatorics problems such as committee selection, solving combination equations, and understanding the difference between permutations and combinations. This detailed guide explains how to form a committee of 5 shareholders from a group of 7 men and 5 women, solve for n in combinatorial equations, and more. Suitable for Grades 9-12.

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