The ordered binomial coefficient I have an arbitrary long list that looks as following: {a,b,c,d,e,f}
Now I would like to create ordered combinations in this list that need to have a length of 3, 4 or 5. To clarify, ordered combinations of length 3 are: {(a,b,c), ((b,c,d), (c,d,e), (d,e,f)} and length 4 are {(a,b,c,d), (b,c,d,e), (c,d,e,f)} and at last we obtain {(a,b,c,d,e), (b,c,d,e,f)} for length 5. The amount of elements in these three created lists is equal to 4+3+2. 
If I wouldn't have set this restriction, all ordered pairs would have the length of n+n-1 ... 1+0 (with n being the length of the original list).
Now the question that I have is:
How could I express these restricted combinations (from length 3,4 or 5) of a list in mathematical terms? And does this have a name? 
Thanks in advance, any help is appreciated.
 A: If your original list has $n$ elements there are $n-k+1$ "ordered combinations" of length $k$ since such a combination can start at any element between the $1$st and the $(n-k)$th. So you have to sum the numbers $n-k+1$ for the values of $k$ in your range. 
If that range consists of consecutive numbers (like your $3,4,5$) that's easy, because the numbers of your combinations will be consecutive integers too, in reverse order. In your example $n=6$ so you want to add 
$$6-3+1,\quad 6-4+1, \quad6-5+1$$ 
which is 
$$ 4,3,2.
$$
Now the sum of the numbers from $1$ to $A$ is
$$
\frac{A(A+1)}{2}
$$
so he sum of the numbers from $B+1$ to $A$ is
$$
\frac{A(A+1)}{2} 
-
\frac{B(B+1)}{2} \ .
$$
In your example you want the sum from $2$ to $4$ so $B=1$ and $A=4$ and you get
$$
\frac{4(4+1)}{2} 
-
\frac{1(1+1)}{2} = 10 - 1 = 9.
$$
PS Your "ordered combinations" don't have any particular name. The only binomial coefficient you need is $\binom{m}{2}$ for finding the two triangular numbers to subtract.
