We need to find the number of functions from the set \( \{1, 2, \dots, 100\} \) to the set \( \{0, 1\} \), such that exactly one of the values in the domain \( \{1, 2, \dots, 100\} \) is mapped to 1, and all other values are mapped to 0.
- First, we select which element from \( \{1, 2, \dots, 98\} \) will be mapped to 1. There are 98 choices for this.
- Then, the remaining 99 elements in the set \( \{1, 2, \dots, 100\} \) must all be mapped to 0. Thus, the total number of functions is \( 98^{99} \).
Final Answer: \( 98^{99} \).
The respective values of \( |\vec{a}| \) and} \( |\vec{b}| \), if given \[ (\vec{a} - \vec{b}) \cdot (\vec{a} + \vec{b}) = 512 \quad \text{and} \quad |\vec{a}| = 3 |\vec{b}|, \] are:
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: