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.

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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 $n$.

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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: $n = 13$.

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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.

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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.