<s>[INST] 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 $7$?
 [/INST] 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 
$$\frac{3\dbinom{7}{5}2! - 2\dbinom{7}{6}1!}{7!}$$
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