Any tips on how to approach the next steps of this problem?
$X$ is a set of $N ≥ 2$ distinct numbers and let $x_1x_2\cdots x_n$ be a permutation of $X$. For $i = 2, 3, . . . , n$ we say that position $i$ in the permutation is a step if $x_{i−1} < x_i$. We consider the position $1$ as a step. What is the expected number of steps in a random permutation of $X$?
First I defined the expectation of $X$ as the sum of the expectation of the random variables for each $x_1+ x_2+\cdots+x_n$ or $\mathbb E[X]= \sum_{i=1}^{n} x_i$.
Now to figure out the expectation of any $x_i$, this is the part I am having trouble with. Each $x_i$ the preceding $x_i-1$ can either be lower or higher so $\mathbb E[ x_i ]= (1\cdot 1/2)-(0\cdot 1/2)$ or $\mathbb E[x_i]= 1/2$. Since we have $n-2$ terms then $\mathbb E[X]= (n-2)\cdot (1/2)$. This reduces so the answer can be shown as $\frac{n-2} 2$. Is this right? If not please let me know what I did wrong and what the right track is, thanks.
Valid approach, except you missed two things: There are $N-1$, not $N-2$, elements in the set $\{2, \ldots, N\}$. (Looks like a fencepost error.) And the problem specifies that the first position always counts as a step. This increases the count for each permutation by $1$, so it also increases the overall expectation by $1$.
So the answer is $1 + \frac{N-1}{2} = \frac{N+1}{2}$.
Another solution:
For each permutation $\sigma \in S_n$, we have a unique permutation which is the reverse of the first, say $R(\sigma)$. Since $N>2$, no permutation is its own reverse; $R(\sigma) \neq \sigma$. And since $R(R(\sigma)) = \sigma$, associating each permutation with its reverse groups the $N!$ permutations into $\frac{N!}{2}$ pairs.
In each pair $\{\sigma, R(\sigma)\}$, each of the $N-1$ pairs of adjacent numbers is increasing in one permutation and decreasing in the other, for a total of $N-1$ steps. The first position in each permutation is also considered a step, so the two permutations together have a total of $2+(N-1)=N+1$ steps.
So plainly the average number of steps is $\frac{N+1}{2}$.