Questions concerning elements in $F = \big\{f: \{1, 2, 3\} \to \{1, 2, 3, 4, 5\}\big\}$. a) Find and simplify the number of functions $f \in F$ so that $f(1) = 4$.
My attempt: there is $1$ choice for $f(1)$, and $5$ choices for $f(2)$ and $5$ choices for $f(3)$, thus $1\cdot 5\cdot 5 = 25$ functions
b) Find and simplify the number of one-to-one functions $f \in F$ so that $f(1) \geq 4$
Attempt: since there are $2$ choices for $f(1)$, I broke it into $2$ cases, one when $f(1) = 5$ and when $f(1) = 4$.
Then there are $(1 \cdot 4 \cdot 3) + (1 \cdot 4 \cdot 3) = 24$ choices.
Is this correct?

Yes, both solutions are correct. Your reasoning for (a) is spot-on.
For (b), you could have simplified your reasoning a little -- there are $2$ choices for $f(1)$, and then $4$ choices for $f(2)$, and $3$ choices for $f(3)$. By the rule of product, you have $2 \cdot 4 \cdot 3 = 24$ choices in total.
