To solve this, we need to evaluate the following two expressions:

1. (a) $\frac{n!}{(n - r)!}$

2. (b) $\frac{n!}{r!(n - r)!}$

where $n = 12$ and $r = 5$.

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Part (a) $\frac{n!}{(n - r)!}$

Given:

• $n = 12$
• $r = 5$

1. First, calculate $n - r$: $n - r = 12 - 5 = 7$

2. Substitute $n = 12$ and $n - r = 7$ into the expression: $\frac{n!}{(n - r)!} = \frac{12!}{7!}$

3. Expand $12!$ and $7!$: $12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7!$

4. Cancel out $7!$ from the numerator and denominator: $\frac{12!}{7!} = 12 \times 11 \times 10 \times 9 \times 8$

5. Calculate this product: $12 \times 11 = 132$ $132 \times 10 = 1320$ $1320 \times 9 = 11880$ $11880 \times 8 = 95040$

So, the answer for part (a) is: $\frac{12!}{7!} = 95040$

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Part (b) $\frac{n!}{r!(n - r)!}$

1. Substitute $n = 12$, $r = 5$, and $n - r = 7$ into the expression: $\frac{n!}{r!(n - r)!} = \frac{12!}{5! \cdot 7!}$

2. Using the values from Part (a), expand $12!$ and cancel out $7!$: $\frac{12!}{5! \cdot 7!} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5! \cdot 7!}$

3. Cancel $7!$ from the numerator and denominator: $= \frac{12 \times 11 \times 10 \times 9 \times 8}{5!}$

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

5. Now we have: $\frac{12 \times 11 \times 10 \times 9 \times 8}{120}$

6. Using the product from Part (a), $12 \times 11 \times 10 \times 9 \times 8 = 95040$: $\frac{95040}{120} = 792$

So, the answer for part (b) is: $\frac{12!}{5! \cdot 7!} = 792$

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Final Answers:

• (a) $\frac{12!}{7!} = 95040$
• (b) $\frac{12!}{5! \cdot 7!} = 792$

Would you like further explanations or have any questions?

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Related Questions:

1. How does factorial growth affect combinations and permutations?
2. How do combinations differ from permutations in calculation?
3. What are the properties of factorials that simplify division?
4. Can you derive the general formula for combinations from factorials?
5. What is the importance of order in permutation problems?

Tip:

Remember, permutations consider order, while combinations do not. This difference is crucial in counting problems!