1. **Problem 12: Recursive definition of factorial** The factorial function $n!$ is defined recursively. We need to identify the correct recursive definition. The correct definition is: $$n! = n \times (n-1)! \text{ for } n \geq 1, \text{ with } 0! = 1$$ Explanation: - Factorial of zero is defined as 1. - For any positive integer $n$, factorial is $n$ times factorial of $(n-1)$. Options C, D, and B are incorrect because: - C) $n! = \frac{n!}{(n-1)!}$ simplifies to $1$, which is not a definition. - D) $n! = (n-1)! \times 2$ is not true for all $n$. - B) $n! = n + (n-1)!$ is addition, not multiplication. 2. **Problem 13: Conclusion of an inductive proof** In mathematical induction, after proving the base case and the inductive step, the conclusion is: $$\text{P(n) is true for all } n \geq \text{base case}$$ Explanation: - Base case shows the statement is true for the initial value. - Inductive step shows if true for $k$, then true for $k+1$. - Together, this proves the statement for all $n$ starting from the base case. Options A, C, and D are incorrect because: - A) Only for $n=k$ is not sufficient. - C) Statement being false contradicts the inductive proof. - D) Hypothesis invalid is incorrect after successful proof. 3. **Problem 14: Calculate $\frac{7!}{5!}$** Recall factorial values: $$7! = 7 \times 6 \times 5!$$ So, $$\frac{7!}{5!} = \frac{7 \times 6 \times 5!}{5!}$$ Cancel $5!$: $$\frac{7 \times 6 \times \cancel{5!}}{\cancel{5!}} = 7 \times 6 = 42$$ **Final answers:** - 12: $n! = n \times (n-1)!$ for $n \geq 1$, with $0! = 1$ - 13: $P(n)$ is true for all $n \geq$ base case - 14: $\frac{7!}{5!} = 42$