Find the number of permutations of the set $\left\{ 1,2,3,4,5,6,7\right\}$ not containing four consecutive elements of ascending order 
Find the number of permutations of the set $\left\{ 1,2,3,4,5,6,7\right\} $ not containing four consecutive elements of ascending order.

My try:
All permutations in cycles are $6!$. 
Let's deal with cases that do not meet the requirements of the task:



*

*When four growing elements are not at the beginning:
 - choice of beginning: ${3 \choose 1}$
 - choice of four numbers ${6 \choose 4}$
 So: $${3 \choose 1}\cdot {6 \choose 4}\cdot 2!=90$$

*When four growing elements are at the beginning:
 $${6 \choose 3}\cdot 3!=120$$
Then we must count cases common to the previous two considerations: $${6 \choose 4}\cdot2!=30$$That is why my answer is: 
$$6!-90-120+30=540$$
However I wrote a Python program to check the number of solutions and he says it's $342$ so I have a mistake.Can you help me and tell me what I'm doing wrong?

 A: Let $A$ be the set of permutations of the sequence $1,2,3,4,5,6,7$ which have $4$ consecutive terms in ascending order.


*

*The number of elements of $A$ whose first $4$ terms are ascending is
$$\binom{7}{4}{\,\cdot\,} 3!=210$$
Explanation:


*

*There are ${\large{\binom{7}{4}}}$ choices for the first $4$ terms.$\\[2pt]$

*There are $3!$ ways to order the $3$ remaining terms.

$\\[2pt]$

*The number of elements of $A$ whose initial block of $4$ consecutive ascending terms are not the first $4$ terms, and which start with the value $1$ is 
$$\binom{6}{3}{\,\cdot\,} 3{\,\cdot\,} 3!=360$$
Explanation:


*

*There are ${\large{\binom{6}{3}}}$ choices for the $3$ terms which follow the value $1$.$\\[2pt]$

*There are $3$ positions where the value $1$ can be placed.$\\[2pt]$

*There are $3!$ ways to order the $3$ remaining terms.

$\\[4pt]$

*The number of elements of $A$ whose initial block of $4$ consecutive ascending terms are not the first $4$ terms, and which start with the value $2$ is 
$$\binom{5}{1}\binom{4}{3}{\,\cdot\,} 3{\,\cdot\,} 2!=120$$
Explanation:


*

*There are ${\large{\binom{5}{1}}}$ choices for the term immediately before the value $2$.$\\[2pt]$

*There are ${\large{\binom{4}{3}}}$ choices for the $3$ terms which follow the value $2$.$\\[2pt]$

*There are $3$ positions where the value $2$ can be placed.$\\[2pt]$

*There are $2!$ ways to order the $2$ remaining terms.

$\\[4pt]$

*The number of elements of $A$ whose initial block of $4$ consecutive ascending terms are not the first $4$ terms, and which start with the value $3$ is 
$$\binom{4}{1}\binom{3}{3}{\,\cdot\,} 3{\,\cdot\,} 2!=24$$
Explanation:


*

*There are ${\large{\binom{4}{1}}}$ choices for the term immediately before the value $3$.$\\[2pt]$

*There are ${\large{\binom{3}{3}}}$ choices for the $3$ terms which follow the value $3$.$\\[2pt]$

*There are $3$ positions where the value $3$ can be placed.$\\[2pt]$

*There are $2!$ ways to order the $2$ remaining terms.

$\\[2pt]$
hence we get
$$|A|=210+360+120+24=714$$
so the number of permutations of the sequence $1,2,3,4,5,6,7$ which do not have $4$ consecutive terms in ascending order is
$$7!-|A|=7!-714=4326$$
A: I'm going to relate to the question how I think you understood it, even though I think quasi's read is much more straightforward. 
(If I'm reading your answer correctly, you're considering cyclic permutations and fixing "the beginning" at $1$.)
You haven't accounted for all the overcounting. Any case with $5$ ascending elements starting not at the beginning has been counted twice in the first step. By your method, there are $2\binom651!=12$ such cases, so we must add $12$.
Then there are the $5$ cases in which exactly $6$ elements appear in order (starting from $1$, with anything other than a $7$ omitted) and the $1$ case in which all appear in order. The former cases were subtracted $3$ times each, then added $2$ times each, so no further correction is needed. The latter case was subtracted $4$ times then added $3$ times, so again no correction is needed.
So the answer should be $552$. I don't know how you got $342$ -- perhaps you can post the Python code? 
