Number of sequences that have at least five consecutive positions in which the numbers are in increasing order? A sequence of numbers is formed from the numbers $1, 2, 3, 4, 5, 6, 7$ where all $7!$ permutations are equally likely. What is the probability that anywhere in the sequence there will be, at least, five consecutive positions in which the numbers are in increasing order?
I approached this problem in the following way, but I am wondering if there is a better way, since my approach is quite complicated.
My Approach: There are three possibilities: a sequence have $7$ consecutive positions in which numbers increase, have $6$ consecutive positions in which numbers increase, and $5$ consecutive positions in which numbers increase.
There is only $1$ sequence that have $7$ consecutive positions. Namely, the sequence $(1,2,3,4,5,6,7)$.
There are $12$ sequences that have $6$ consecutive positions. Namely, we choose $1$ number from $(1,2,3,4,5,6,7)$, and move it to either sides. As an illustration, if we choose $3$, then we can get $(3,1,2,4,5,6,7)$ or $(1,2,4,5,6,7,3)$.
Now consider when there are $5$ consecutive positions in which numbers increase. We choose $2$ numbers that are not in the increasing subsequence. 
If $1$ and $7$ are not chosen, we can place them in front of the subsequence, of after. For example, if we choose $(2,5)$, then we will have $(2,5,1,3,4,6,7)$,$(5,2,1,3,4,6,7)$, $(1,3,4,6,7,2,5)$ and $(1,3,4,6,7,5,2)$. This is $\binom{5}{2}\times4$.
Then I'm not sure how to proceed when we choose $1$ and/or $7$?

Both myself and @N.F.Taussig used the following approach, although I'd like to see if it could be generalised to increasing runs of arbitrary length.
Define set $S_{i,j}$ as the set of permutations of $[7]$ with an increasing run between position $i$ and $j$ inclusive. Then by inclusion-exclusion the desired success count is
$$\begin{align}&(|S_{1,5}|+|S_{2,6}|+|S_{3,7}|) -\\ (|S_{1,5}\cap S_{2,6}| + |S_{1,5}&\cap S_{3,7}|+|S_{2,6}\cap S_{3,7}|)+ |S_{1,5}\cap S_{2,6}\cap S_{3,7}|\tag{1}\end{align}$$
Clearly
$$|S_{1,5}|=|S_{2,6}|=|S_{3,7}|=\binom{7}{5}2!\tag{2}$$
since we choose $5$ of the $7$ numbers to go in increasing order in positions $1$ to $5$, $2$ to $6$ or $3$ to $7$ and the remaining $2$ numbers can go in the remaining $2$ spots in $2!$ ways.
Also
$$S_{1,5}\cap S_{2,6}=S_{1,6}$$
$$\implies |S_{1,5}\cap S_{2,6}|=|S_{1,6}|=\binom{7}{6}1!\tag{3}$$
and
$$S_{1,5}\cap S_{3,7}=S_{1,7}$$
$$\implies |S_{1,5}\cap S_{3,7}|=|S_{1,7}|=\binom{7}{7}0!\tag{4}$$
and
$$S_{2,6}\cap S_{3,7}=S_{2,7}$$
$$\implies |S_{2,6}\cap S_{3,7}|=|S_{2,7}|=\binom{7}{6}1!\tag{5}$$
and
$$S_{1,5}\cap S_{2,6}\cap S_{3,7}=S_{1,7}$$
$$\implies |S_{1,5}\cap S_{2,6}\cap S_{3,7}|=|S_{1,7}|=\binom{7}{7}0!\tag{6}$$
using similar reasoning to $(2)$ in each case.
Putting the results of $(2)$, $(3)$, $(4)$, $(5)$ and $(6)$ into $(1)$ gives:
$$\text{success count}=3\binom{7}{5}2!-\left(2\binom{7}{6}1!+\binom{7}{7}0!\right)+\binom{7}{7}0!=112$$
Then since there are $7!$ permutations the desired probability is:

$$\text{probability of an increasing run of length $\ge 5$}=\frac{112}{7!}=\frac{1}{45}\tag{Answer}$$

