Here is another way of visualising Moron's answer:
Let $S$ be the first set, and $T$ be the second set. Now, imagine the elements of $S$ are listed, horizontally, in their order; we will not disturb their order, but will only "place" the elements of $T$ among them.
The possible locations for any element of $T$ are (i) before the first elements of $S$; (ii) between two elements of $S$; and (iii) after the last element of $S$. If $S$ has $s$ elements, then this gives $s+1$ possible locations. We can put more than one element of $T$ in each location, though. For example, if $S=\{1,2,3\}$ and $T=\{a,b,c,d\}$, then we can place elements of $T$ either before $1$, between $1$ and $2$, between $2$ and $3$, or after $3$ (four locations); and we could place one element before $1$, none between $1$ and $2$, two between $2$ and $3$, and one after $3$. But note that once we decide where to place the elements of $T$, the order they will appear on is completely determined. If we go the way I just described, you would end up with $a,1,2,b, c, 3, d$, and that is the only way to place elements of $T$ as described while preserving the order.
If we are placing $t$ elements, then, we just need to "select" $t$ locations from $s+1$ possibilities. The order in which we pick them doesn't matter, because in the end we will just put the elements of $T$ in their appropriate order in those locations. And we can select the same location more than once. So we need to compute "combinations with repetitions" (order does not matter, repetitions allowed). The formula for making $n$ selections, with repetitions allowed but where order does not matter, from among $m$ possibilities is $\binom{n+m-1}{n}$ (see for example here) so here we have $n=t$ and $m=s+1$, giving $\binom{t+s}{t} = \binom{t+s}{s}$ possibilities.