$24!$ counts the number of ways that you can rearrange 24 distinct objects.
The expression you give counts the number of ways that you can select a subset of your 24 distinct objects, and then rearrange them into a new order.
For example, with 3 objects {1,2,3} you could choose any of the following orderings:
- (1) (2) (3)
- (1,2) (1,3) (2,1) (2,3) (3,1) (3,2)
- (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,1,2) (3,2,1)
The total number of ways is
(# of ways to select one object) + (# of ways to select two objects) + (# of ways to select 3 objects)
which is equal to
$\frac{3!}{2!} + \frac{3!}{1!} + \frac{3!}{0!}$
or equivalently
$\sum_{m=1}^2 \frac{3!}{(3-m)!} + 3!$
which is the expression you give in your question, but for the case of 3 objects rather than 24. In fact you could write it even more succinctly:
$\sum_{m=1}^3 \frac{3!}{(3-m)!}$
Moving slightly beyond your question, I would argue that there is one more way to select a subset of those objects, namely selecting none of them at all, in which case there are
$\sum_{m=0}^3 \frac{3!}{(3-m)!}$
In addition, I would note that due to a symmetry between $m$ and $3-m$, you can write this as
$\sum_{m=0}^3 \frac{3!}{m!}$
Finally, let's generalize to the case of $n$ objects, in which case the number of ways of selecting a subset and rearranging them into a new order is
$a(n) = \sum_{m=0}^n \frac{n!}{m!}$
There is an (almost) closed-form formula for this: if $n>0$ then
$a(n) = \lfloor en! \rfloor$