Consider the sets $A = \left\{0, 1, 2, 3,4\right\}$ and $B = \left\{1,2,3,4,5,6,7,8,9,10\right\}$. Let $F$ be the set of functions whose domain is $A$ and whose codomain $B$. If one randomly chooses an $ f$ function of $F$, the probability of $f$ being strictly increasing or being injective is?
Attemp: I feel like there is a typo as aren't all strictly increasing functions also injective since both injective and strictly increasing functions have the condition that $\displaystyle \forall a,b\in X,\;\;f(a)=f(b)\Rightarrow a=b$.
(It's a matter of a list)
So all we have to do is count the number of injective functions for the function to be strictly increasing or injective. That's simply $ 10P5 = 30240$ as we can imagine giving a element of $B$ to a element of $A$.
Correct?
You're right, every strictly increasing function is also injective. If $S_1$ is the set of all strictly increasing functions and $S_2$ is the set of all injective functions, then $S_1 \subset S_2$ and thus $|S_1 \cup S_2| = |S_1|+|S_2|-|S_1\cap S_2| = |S_1|+|S_2|-|S_1| = |S_2|$, so your solution is correct.