Revisit : $20\choose 5$ subsets without 3,4 or 5 consecutive numbers Addendum-2 just added to my question.

Addendum just added to my question.

$\underline{\textbf{Overview}}$
This is a self-answer question of
this original question.
I strongly suspect that the original question will soon be closed and then deleted.

I’m trying to get the amount of combinations of 5 numbers from one to twenty
without duplicates and without 3,4,and 5 consecutive running numbers.

$\underline{\textbf{Clarification}}$
Let $N = \{1,2,\cdots, 20\}$.  
How many distinct subsets of $N$ are there where:

*

*The subset has exactly $5$ elements.

*The subset does not contain $3$ consecutive elements. 
Here, consecutive elements are elements $(k), (k+1), (k+2).$
For example, both of the following sets are satisfactory:

*

*$\{1, 2, 4, 5, 7\}$

*$\{1, 3, 5, 7, 9\}$.

Further, each of the following sets are unsatisfactory:

*

*$\{1, 2, 3, 14, 18\}$

*$\{2, 3, 4, 5, 17\}$

*$\{8, 9, 10, 11, 12\}$.

$\underline{\textbf{My Background}}$
About $50$ years ago I took a Probability course in college and did ok.  I have
forgotten much of the theory, and usually rely exclusively on intuition to attack
Probability (or Combinatorics) problems.
If relevant, some decades ago I survived but have forgotten much of:

*

*"Real Analysis : Volume 1 : 2nd Ed." (Apostol, 1966).


*The first $(2/3)$ of "Elementary Number Theory" (Uspensky and Heaslett, 1938)
[through quadratic reciprocity].
$\underline{\textbf{Problem Relevance}}$
In my experience, there are three typical approaches to this type of problem:

*

*The Direct Approach

*Recursion

*Inclusion-Exclusion

This particular problem interested me, because of the challenge involved in providing three distinct solutions, one for each of the above approaches.  However, exploring Inclusion-Exclusion,
I concluded that the math involved was too ugly to be reasonably feasible.
However, I was able to find two distinct Direct Approaches to offer.
$\underline{\textbf{My Work}}$
See my self - answers.  
For clarity, I have provided a separate answer for :

*

*A Direct Approach

*An Alternate Direct Approach

*Recursion


Addendum
Given the answers provided by others, it seems to me that the one pending challenge is to find some elegant solution that is based primarily on Inclusion-Exclusion.
I would be very interested if someone could present such a solution.
Edit
Mike Earnest added an Inclusion-Exclusion response to his answer.

Addendum-2
Finally conquered my own private Inclusion-Exclusion challenge for this problem.  Just added a separate Inclusion-Exclusion answer.
 A: According to the beautiful explanation , these $5$ numbers can consists of two ways suchthat

*

*No consecutive numbers


*Only one pair of consecutive numbers
So ,

*

*For no consecutive numbers :

Lets represent the numbers by letters such that the selected numbers is represened by $\color{blue}{S}$ , and non-selecteds by $\color{red}{S}$.
Now , we must have $15\color{red}{S'}$ and $5\color{blue}{S'}$. Now ,we will distribute these $5$ blue $S'$ among and ends of $15\color{red}{S'}$ . We can do it by $C(16,5)$ ways.For example , one of the distribution is $$\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}$$
Now , think that $$1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20=\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}$$
So , we select the numbers $\{1,5,8,10,12\}$

*

*For only one pair of consecutive numbers :

Select one of the possible place among $16$ gaps (ends and between the red letters), and place two blue letters in that place. We can do it by $C(16,1)$ ways.By using the same logic , select $3$ place for the remaining letters among $15$ suitable places by $C(15,3)$.For example , $$\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}$$
Above ,we select $\{1,5,8,11,12\}$
So , the answer is $$C(16,5) + C(16,1) \times C(15,3) = 4368+16 \times455 =11648$$
$\mathbf{EDITION}$: We can also have two separate consecutive numbers such as $\{1,2,5,7,8\}$
So , select $2$ places among $16$ suitable places by $C(16,2)$ to place double blue letters and select one place for the remaining by $C(14,1)$.
For example , $$\color{blue}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{red}{S}\color{blue}{S}\color{blue}{S}$$
We select $\{1,2,10,19,20\}$
So , the answer is $$C(16,5) + [C(16,1) \times C(15,3) ]+ [C(16,2) \times C(14,1)] = 4368+[16 \times455] +[120 \times 14] = \color{blue}{13,328}$$
A: $\underline{\textbf{A Direct Approach}}$
For any set $T$ with a finite number of elements, let $|T|$ denote the number of
elements in $T$.
Let $N$ denote the set $\{1, 2, \cdots, 20\}.$ 
Let $A$ denote the following collection of subsets of $N$: 
$\{S \subseteq N ~: ~S ~\text{has exactly} ~5 ~\text{elements}\}.$
For $~k \in \{1, 2, \cdots, 18\},~$ 
let $B_k$ denote the collection of all subsets $S$ in $A$ where:

*

*$S~$ specifically contains the elements $~(k), (k+1),~$ and $~(k + 2)$.


*When $k > 1$, the subset $S$ does not contain the element $(k-1)$.
This means that if $S$ is in $B_k$, $S$ contains a group of (at least)
$3$ consecutive elements
such that the lowest element in this group is the element
$k$.  This implies that $B_1, B_2, B_3, \cdots, B_{18}$ are all disjoint sets.
This implies that the desired computation is
$$ |A| - \left[\sum_{k=1}^{18} |B_k|\right]. $$

To enumerate $|B_2|,~$ for any subset $S$ in $B_2$:

*

*$S$ does not contain the element $(1)$.


*$S$ does contain the elements $~(2), (3),~$ and $~(4)$.


*$S$ contains any $2$ other elements from $\{5, 6, \cdots, 20\}$.
Therefore
$$|B_2| = \binom{16}{2}.\tag1 $$
Similarly, for $k \in \{3, 4, \cdots, 16\}$, 
for any subset $S$ in $B_k$:

*

*$S$ does not contain the element $(k-1)$.


*$S$ does contain the elements $~(k), (k+1),~$ and $~(k+2)$.


*$S$ contains any $2$ other elements from the $16$ remaining elements in $N$.
Therefore
$$|B_3| = |B_4| = \cdots = |B_{16}| = \binom{16}{2}.\tag2 $$

$\underline{\text{Special Handling}}$
For $~k \in \{1, 17, 18\}$ the computation of
$|B_k|$ will need special handling.
For any subset $S$ in $B_1$:

*

*$S$ does contain the elements $~(1), (2),~$ and $~(3)$.


*$S$ contains any $2$ other elements from $\{4, 5, \cdots, 20\}$.
Therefore,
$$|B_1| = \binom{17}{2}. \tag3 $$
For any subset $S$ in $B_{17}$:

*

*$S$ does contain the elements $~(17), (18),~$ and $~(19)$.


*$S$ contains any $2$ other elements from either $\{20\}$ or
$\{1, 2, \cdots, 15\}$.
Therefore,
$$|B_{17}| = |B_2| = \binom{16}{2}. \tag4 $$
For any subset $S$ in $B_{18}$:

*

*$S$ does contain the elements $~(18), (19),~$ and $~(20)$.


*$S$ contains any $2$ other elements from
$\{1, 2, \cdots, 16\}$.
Therefore,
$$|B_{18}| = |B_2| = \binom{16}{2}. \tag5 $$

$\underline{\text{Final Computation}}$
Putting the computations in (1), (2), (3), (4), and (5) above all together,
$$|A| - \left[\sum_{k=1}^{18} |B_k|\right]$$
$$ = \binom{20}{5} - \left[17 \times \binom{16}{2}\right] - \binom{17}{2}$$
$$ = 15504 - \left[17 \times 120\right] - 136 = 13328.$$
A: $\underline{\textbf{An Alternate Direct Approach}}$
For a set $T$ with a finite number of elements, 
let $|T|$ denote the number of elements in $S$.
Let $N$ denote the set $\{1, 2, \cdots, 20\}.$
Let $A$ denote the following collection of subsets of $N$: 
$\{S \subseteq N ~: ~S ~\text{has exactly} ~5 ~\text{elements}\}.$
Let $B$ denote the following collection of subsets $S$ that are elements in $A$:
$S$ has $3$ consecutive elements but does not have $4$ consecutive elements.
Let $C$ denote the following collection of subsets $S$ that are elements in $A$:
$S$ has $4$ consecutive elements but does not have $5$ consecutive elements.
Let $D$ denote the following collection of subsets $S$ that are elements in $A$:
$S$ has $5$ consecutive elements.
Let $a,b,c,d$ denote $|A|, |B|, |C|, |D|,$ respectively $\implies$

*

*$a = \displaystyle \binom{20}{5} = 15504.$

*The desired computation is $a - (b + c + d).$

$\underline{\text{To Compute} ~b}$
For $k \in \{1, 2, \cdots, 18\},$ 

*

*let $B_k$ denote the collection of subsets $S$ where 
$\{S \in B ~: ~S ~\text{contains the elements} ~k, k+1, ~\text{and} ~k+2\}.$

*let $b_k$ denote $|B_k|$.

This implies that $\displaystyle ~b = \sum_{k=1}^{18} b_k$.
$\displaystyle b_1 = \binom{16}{2}$. 
This is because any set $S \in B_1$ contains the elements 
$(1), (2),$ and $(3)$ 
does not contain the element $(4)$, 
and contains $2$ elements from $\{5, 6, \cdots, 20\}$.
Similarly, $\displaystyle ~b_{18} = \binom{16}{2}.$ 
$\displaystyle b_2 = \binom{15}{2}$. 
This is because any set $S \in B_2$ contains the elements 
$(2), (3),$ and $(4)$ 
does not contain either of the elements $(1)$ or $(5)$, 
and contains $2$ elements from $\{6, 7, \cdots, 20\}$.
Similarly, for $\displaystyle ~k \in \{3, 4, \cdots, 17\}, ~b_k = \binom{15}{2}$.
Therefore,
$$b = \left[2 \times \binom{16}{2}\right] + \left[16 \times \binom{15}{2}\right] 
= 240 + 1680 = 1920.$$

$\underline{\text{To Compute} ~c}$
For $k \in \{1, 2, \cdots, 17\},$ 

*

*let $C_k$ denote the collection of subsets $S$ where 
$\{S \in C ~: ~S ~\text{contains the elements} ~k, k+1, k+2, ~\text{and} ~k+3\}.$

*let $c_k$ denote $|C_k|$.

This implies that $\displaystyle ~c = \sum_{k=1}^{17} c_k$.
$\displaystyle c_1 = \binom{15}{1}$. 
This is because any set $S \in C_1$ contains the elements 
$(1), (2), (3)$ and $(4)$ 
does not contain the element $(5)$, 
and contains $1$ element from $\{6, 7, \cdots, 20\}$.
Similarly, $\displaystyle ~c_{17} = \binom{15}{1}.$ 
$\displaystyle c_2 = \binom{14}{1}$. 
This is because any set $S \in C_2$ contains the elements 
$(2), (3), (4)$ and $(5)$ 
does not contain either of the elements $(1)$ or $(6)$, 
and contains $1$ element from $\{7, 8, \cdots, 20\}$.
Similarly, for $\displaystyle ~k \in \{3, 4, \cdots, 16\}, ~b_k = \binom{14}{1}$.
Therefore,
$$c = \left[2 \times \binom{15}{1}\right] + \left[15 \times \binom{14}{1}\right] 
 = 30 + 210 = 240.$$

$\underline{\text{To Compute} ~d}$
For $S \in D$, there are $(16)$ possible choices for the lowest element in $S$.
$$d = 16.$$

$\underline{\text{Final Computation}}$
$$a - (b + c + d) = 15504 - (1920 + 240 + 16) = 13328.$$
