For the given values of n and​ r, evaluate ​(a) StartFraction n exclamation mark Over left parenthesis n minus r right parenthesis exclamation mark EndFraction
 and ​(b) StartFraction n exclamation mark Over r exclamation mark left parenthesis n minus r right parenthesis exclamation mark EndFraction
 for nequals12 and requals5.
Question content area bottom
Part 1
​(a) When nequals12 and requals5​, StartFraction n exclamation mark Over left parenthesis n minus r right parenthesis exclamation mark EndFraction
equals
  
enter your response here.
​(Simplify your answer. Type an integer or a simplified​ fraction.)

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 \(r = 5\).

---

### 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:
\[
\frac{12!}{7!} = 95040
\]

---

### 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:
\[
\frac{12!}{5! \cdot 7!} = 792
\]

---

### Final Answers:
- **(a)** \(\frac{12!}{7!} = 95040\)
- **(b)** \(\frac{12!}{5! \cdot 7!} = 792\)

Would you like further explanations or have any questions?

---

#### 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!

For the given values of n and r, evaluate (a) \(\frac{n!}{(n - r)!}\) and (b) \(\frac{n!}{r!(n - r)!}\) for \(n = 12\) and \(r = 5\).

This solution demonstrates how to calculate factorial expressions for permutations and combinations. For n = 12 and r = 5, evaluate both \(\frac{12!}{7!}\) and \(\frac{12!}{5! \cdot 7!}\), simplifying step-by-step to 95040 and 792, respectively. Ideal for high school math students exploring counting principles.

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