This is a variation of the combinatorial problem considered in section 5 of <A HREF="http://www.brynmawr.edu/math/people/anmyers/PAPERS/Coupons.pdf">
Some new aspects of the coupon collector’s problem</A> (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t-k$ classes, $k=0,1,\ldots s-t-1$, no class having fewer than two elements, which can be chosen in $\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix}
n-t\\
s-t-k
\end{Bmatrix}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{(s-t)!s^n} {n\choose t}\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix}
n-t\\
s-t-k
\end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix}
n\\
k
\end{Bmatrix}_2$ is
$$\sum_{n\geq 0}\begin{Bmatrix}
n\\
k
\end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$