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 $7$?
 A: 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:

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

A: This solution does not differ in any essential way from that of N. Shales.  I am posting this here so N. Shales can compare our approaches.
Since the sequence contains seven numbers, any block of five consecutive increasing numbers must start in the first, second, or third positions.  Let $A_1$, $A_2$, and $A_3$ denote, respectively, the set of sequences containing five consecutive increasing numbers that begin in the first, second, and third positions.
By the Inclusion-Exclusion Principle, the number of sequences containing a block of at least five consecutive increasing numbers is 
$$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$:  There are $\binom{7}{5}$ ways to select five of the seven numbers to be in a block of five consecutive increasing numbers and one way to arrange the numbers within that block so that they are increasing.  There are $2!$ ways to arrange the remaining two numbers in the remaining two positions.  Hence, 
$$|A_1| = \binom{7}{5}2!$$
By symmetry, $|A_1| = |A_2| = |A_3|$.
$|A_1 \cap A_2|$:  If both the first five and second five numbers are increasing, then the first six numbers must form an increasing sequence.  There are $\binom{7}{6}$ ways to select six of the seven numbers to be in the block of six consecutive increasing numbers and one way to arrange the numbers within the block so that they are increasing.  There is one way to place the remaining number in the remaining position.  Hence, 
$$|A_1 \cap A_2| = \binom{7}{6}1!$$
By symmetry, $|A_1 \cap A_2| = |A_2 \cap A_3|$.
$|A_1 \cap A_3|$:  If the first five and last five of the seven numbers are increasing, then all seven numbers must be increasing.  There is only one way to arrange all seven numbers in an increasing sequence.  Hence, 
$$|A_1 \cap A_3| = \binom{7}{7}$$
$|A_1 \cap A_2 \cap A_3|$:  There is only one way for all seven numbers to be arranged in increasing order.  Hence, 
$$|A_1 \cap A_2 \cap A_3| = \binom{7}{7}$$
Therefore, 
$$|A_1 \cup A_2 \cup A_3| = 3\binom{7}{5}2! - 2\binom{7}{6}1! - \binom{7}{7} + \binom{7}{7} = 3\binom{7}{5}2! - 2\binom{7}{6}$$
Hence, the probability that a sequence of seven numbers contains a block of at least five consecutive increasing numbers is 
$$\frac{3\dbinom{7}{5}2! - 2\dbinom{7}{6}1!}{7!}$$
A: ${}_7 C_2 = 21$, are the number of ways of choosing two out of the seven numbers. We pull them out. The remaining five retain their increasing order. So now find the number of ways you can replace those two values at either the beginning and/or end of the sequence, and multiply this result by 21. If they happen to be replaced where they started then the length of the increasing sub-sequence grows, but this is allowed anyway. Two elements can be replaced in the sequence in 6 ordered ways, by my count. That is $2!=2$ orders times 3 ways of splitting them up.  In total then I figure there are 126 ways of creating sequences with 5 or more increasing elements anywhere in the sequence.  Divide by your $7!=5040$ permutations for a probability.
