1. **Problem:** Show that the function $f:\mathbb{Z} \times \mathbb{Z} \to \mathbb{Z}$ defined by $f(m,n) = m + n$ is onto. 2. **Formula and Explanation:** A function $f$ is onto (surjective) if for every element $y$ in the codomain, there exists at least one element $(m,n)$ in the domain such that $f(m,n) = y$. 3. **Work:** Given any integer $y \in \mathbb{Z}$, choose $m = y$ and $n = 0$. Then $$f(m,n) = f(y,0) = y + 0 = y.$$ This shows for every $y$ in the codomain, there is a pair $(m,n)$ in the domain mapping to it. 4. **Conclusion:** Therefore, $f$ is onto. --- 1. **Problem:** Show that if $f: B \to C$ and $g: A \to B$ are one-to-one (injective), then the composite function $f \circ g: A \to C$ is one-to-one. 2. **Formula and Explanation:** A function is one-to-one if $f(x_1) = f(x_2)$ implies $x_1 = x_2$. 3. **Work:** Suppose $(f \circ g)(a_1) = (f \circ g)(a_2)$ for $a_1,a_2 \in A$. Then $$f(g(a_1)) = f(g(a_2)).$$ Since $f$ is one-to-one, this implies $$g(a_1) = g(a_2).$$ Since $g$ is one-to-one, this implies $$a_1 = a_2.$$ 4. **Conclusion:** Hence, $f \circ g$ is one-to-one. --- 1. **Problem:** How many different 7-place license plates are possible if the first 3 places are vowels and the last 4 are numbers, with repetition allowed? 2. **Explanation:** There are 5 vowels (A, E, I, O, U) and 10 digits (0-9). 3. **Work:** Number of ways to fill first 3 places with vowels: $$5^3 = 125.$$ Number of ways to fill last 4 places with digits: $$10^4 = 10000.$$ Total number of license plates: $$5^3 \times 10^4 = 125 \times 10000 = 1,250,000.$$ 4. **Conclusion:** There are 1,250,000 possible license plates. --- 1. **Problem:** Prove that $P(n,r) = \frac{n!}{(n-r)!}$, where $P(n,r)$ is the number of permutations of $n$ objects taken $r$ at a time. 2. **Explanation:** Permutations count ordered arrangements. 3. **Work:** To form an ordered arrangement of $r$ objects from $n$: - Choose the first object: $n$ choices - Choose the second object: $n-1$ choices - ... - Choose the $r$-th object: $n-r+1$ choices Multiply: $$P(n,r) = n \times (n-1) \times \cdots \times (n-r+1).$$ This product equals $$\frac{n!}{(n-r)!}$$ because $$n! = n \times (n-1) \times \cdots \times (n-r+1) \times (n-r)!.$$ 4. **Conclusion:** Thus, $P(n,r) = \frac{n!}{(n-r)!}$.