Combinations without repetition and with limit on consecutives I see from Wikipedia that Binomial coefficient finds the number of k-combinations from a given set of n elements.
In this figure we have 10 3-element subsets of a 5-element set:

In my problem I have a restriction: I can't have more than M consecutive elements in my subsets. For example, in the previous figure if M=2 than I should exclude the 1st row, the 7th one and the last one. I'm interested in the resulting number of rows (in the example, 10-3=7 rows).
How can I count the correct number of rows taking into account this restriction? I imagine something like this:
nRows = C(n,k) - nRowsWithConsecutiveElements(n,k,M)

 A: Starting from
$$
\prod\limits_{1 \le \,j\, \le \,n} {\left( {1 + x} \right)}  = \left( {1 + x} \right)^n  = \sum\limits_{0\, \le \,k\, \le \,n} {\binom{n}{k}x^{\,k} } 
$$
we know what the binomial coefficient is indicating.
Let's consider then
$$
\prod\limits_{1\, \le \,j\, \le \,n} {\left( {1 + x\,y^j } \right)}
  = \sum\limits_{0\, \le \,k\, \le \,n} {\;\sum\limits_{0\, \le \,\binom{k+1}{2}\, \le \,j\, \le \,k\,n - \binom{k}{2}
 \, \le \,\binom{n+1}{2}\,} {c_{\,k,\,j} \,x^{\,k} y^{\,j} } } 
$$
where $c_{\,k,\,j}$ represents the number of partitions of $j$ into exactly $k$ different parts of size at most $n$, i.e.
$$
c_{\,k,\,j}  = \left| {\left\{ {\left\{ {j_1 ,j_2 , \cdots ,j_k } \right\}:\left\{ \matrix{
  1 \le j_1  < j_2  <  \cdots  < j_k  \le n \hfill \cr 
  j_1  + j_2  +  \cdots  + j_k  = j \hfill \cr}  \right.} \right\}} \right|
$$
Clearly, putting $y=1$, we will have
$$
\sum\limits_{0\, \le \,j\, \le \,\binom{n+1}{2} } {c_{\,k,\,j} }  = \binom{n}{k}
$$
Now, among the combinations of $k$ integers $j_l$ summing to $j$ we want to count 
those which include $m$ consecutive integers.
If at least $m$ of the $k$ parts are consecutives, then the inequalities can be reformulated as
$$
\eqalign{
  & 1 \le j_1  <  \cdots  < j_l  < \underbrace {j_l  + 1 < j_l  + 2 <  \cdots  < j_l  + m}_m < j_{l + m + 1}  <  \cdots  < j_k  \le n  \cr 
  & \quad \quad \quad \quad  \Downarrow   \cr 
  & 1 \le j_1  <  \cdots  < j_l  < j_{l + 1}  <  \cdots  < j_{k - m}  \le n - m \cr} 
$$
while for the sum we have
$$
\eqalign{
  & j_1  +  \cdots  + j_l  + \underbrace {j_l  + 1 + j_l  + 2 +  \cdots  + j_l  + m}_m + j_{l + m + 1}  +  \cdots  + j_k  = j  \cr 
  & \quad \quad \quad \quad  \Downarrow   \cr 
  & j_1  +  \cdots  + j_l  + j_{l + 1}  +  \cdots  + j_{k - m}
  = j - \binom{m+1}{2} - m\,j_l \quad \left| \matrix{
  \;0\, \le j_l  \le n - m \hfill \cr 
  \;\binom{k+1}{2}\, \le j \le kn - \binom{k}{2}
  \hfill \cr}  \right. \cr} 
$$
So we have the same scheme as above, with the paramaters changed to
$$
\left\{ {\matrix{
   n \hfill &  \to  \hfill & {n - m} \hfill  \cr 
   k \hfill &  \to  \hfill & {k - m} \hfill  \cr 
   j \hfill &  \to  \hfill & {j^{\, * }  = j - \binom{m+1}{2}
 - l\,m\quad \left| \matrix{
  \;0\, \le l \le n - m \hfill \cr 
  \;0\, \le \,\binom{k+1}{2} \, \le \,j\, \le \,k\,n - \binom{k}{2} \, \le \,\binom{n+1}{2} \, \hfill \cr}  \right.} \hfill  \cr 
 } } \right.
$$
where of course $j^{\, * } $ shall remain non-negative.
In your example for instance, we have
$$
\left\{ \matrix{
  n = 5\quad k = 3\quad m = 3\quad 6 \le j \le 12 \hfill \cr 
  n - m = 2\quad k - m = 0\quad 0 \le l \le 2 \hfill \cr 
  j^{\, * }  = j - \binom{m+1}{2} - l\,m = j - 6 - 3l \hfill \cr}  \right.
$$
so that among the $\binom{5}{3}=10$ 3-subsets $\{j_1,j_2,j_3\}$ of $\{1,2,\cdots,5\}$, the number of those having three
consecutive digits corresponds to the sum of all the $c_{k-m,\, j^{\,*}}$ in
$$
\eqalign{
  & \left[ {x^{\,k - m} } \right]\prod\limits_{1\, \le \,j\, \le \,n - m} {\left( {1 + x\,y^{\,j^{\, * } } } \right)}
  = \left[ {x^{\,0} } \right]\prod\limits_{1\, \le \,j\, \le \,2} {\left( {1 + x\,y^{\,j^{\, * } } } \right)}  =   \cr 
  &  = x^{\,0} \sum\limits_{0\,\, \le \,j\, \le \,0\,} {c_{\,0,\,j^{\, * } } \,y^{\,j^{\, * } } }  = 1\,y^{\,0} \quad  \to \quad j^{\, * }  = 0 \cr} 
$$
i.e. to the three solutions given by $j=6,9,12$

Finally let's evidentiate the important fact that the product taken into consideration can be written as
$$
\eqalign{
  & \prod\limits_{1\, \le \,j\, \le \,n} {\left( {1 + x\,y^j } \right)}  =   \cr 
  &  = {1 \over {1 + x}}\prod\limits_{0\, \le \,j\, \le \,n} {\left( {1 + x\,y^j } \right)}  = {1 \over {1 + x}}\left( { - x\,;y} \right)_{\,n + 1}  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,n - 1} {\left( {1 + \left( {xy} \right)\,y^j } \right)}  = \left( { - xy\,;y} \right)_{\,n}  \cr} 
$$
where
$$
\left( {a\,;q} \right)_{\,n} 
$$
indicates the q-Pochhammer symbol.
Consequently, the $c_{\, k \, j}$'s  can be expressed in term of q-Binomial coefficients.
