A probability problem by VANSH SARDANA

There should be $510$ such functions. There are $10^{10}$ total functions, so the probability is $\frac{51}{10^9},$ and the number of factors of $51$ is $\fbox{4}.$

To count these functions, first consider the minimum value $m$ of such a function on $\{2,\ldots,9\}.$ If it's not $1,$ then the function is $1$ at one of the endpoints, and $m$ somewhere in the middle. In between, the function is strictly larger than $m$ (and there must be some $x$ values in between, so that $m$ can be a local minimum), so it must attain a local maximum somewhere in between, which is impossible. Hence the local minimum is attained at $(x,1)$ for some $x \in \{2, \ldots, 9\}.$

Now I claim that there is a one-to-one correspondence between such functions and choices of a nonempty, non-full subset of $\{2, \ldots, 10\}.$ Given such a subset, let $f$ be the function that takes on the values in the set in descending order, then takes on $1,$ then takes on the values in the complement of the set in ascending order. For example, if we choose the subset $\{4,6,9\},$ then $f$ is the function that takes on the values $9,6,4,1,2,3,5,7,8,10$ at the $x$ -values $1,2,3,\ldots,10$ respectively.

Given the discussion about the local minimum above, it's easy to check that this is a bijection. There are $2^9-2 = 510$ such subsets, so the result follows.