Ways to select $m$ disjoint groups of consecutive elements from an $n$-element set In how many ways can you select $m$ disjoint groups of consecutive elements (of any size $\geq 1$) from an $n$-element set?
As an example, I'm thinking of the following case: imagine $S=\{1,3,5,7,9,11,15\}$ so that $n=|S|=7$. The ways to select $1$ (disjoint has no sense here) group of consecutive elements from this set are $\binom{7}{2} + \binom{7}{1}$: all 1 element subsets and all subsets which are defined by the two ends for the consecutive elements to be included (e.g., the choice of $3$ and $11$ gives the set $\{3,5,7,9,11\}$).
 A: In your general case, there are $m$ beginnings and $m$ ends applied to the ordered set of elements
so for example if $m=2$ you get $.(.).(.).$ with the possibility of putting elements in the $.$s
but each $($ must be followed by an element so actually you have $.(x.).(x.).$ reducing the number of free elements to place down to $n-m$. If $m=2$, say $a,b,c$, and $n=3$ then you can position one free element in the $.$s and so, since there are $5$ .s, there are $5$ possibilities:

*

*$a(b)(c)$ where the free element is $a$

*$(ab)(c)$ where the free element is $b$

*$(a)b(c)$ where the free element is $b$

*$(a)(bc)$ where the free element is $c$

*$(a)(b)c$ where the free element is $c$
So you need to choose the positions of the $2m$ things fixed by the beginnings and ends and the $n-m$ "free" elements, which you can do in ${2m+n-m \choose 2m} = {n+m \choose 2m}$ ways.
For your example of $n=7$ and $m=1$ you would get ${7+1\choose 2\times 1}={8 \choose 2}=28$ which is equal to your $\binom{7}{2} + \binom{7}{1}$.
