Probability of a certain dice roll sum disregarding lowest rolls The number of ways to obtain a total of $p$ in $n$ rolls of $s$-sided dice is:
$$c=\sum_{k=0}^{\lfloor(p-n)/s\rfloor}(-1)^k\binom{n}k\binom{p-sk-1}{n-1}\;.$$
What I'm interested in is making the $n$ rolls, but then disregarding the lowest $m$ rolls. E.g. out of five rolls $1, 5, 3, 5, 6$, disregard the lowest two - the $1$ and $3$.
I want to know the number of ways to obtain $p$ as the sum of the remaining $n-m$ rolls. (In my example, the sum was $5+5+6=16$, with of course $n=5, m=2$)
 A: This is a brute force approach, resulting in very long expressions and summations. I am not sure if this is helpful, but I guess, it is a start.
Let $c(n,p,s)$ denote the number of ways to get a sum of $p$ in $n$ rolls of $s$-sided dice. If it is given that the values of all the rolls are at least $r$ and the value $r$ occurs exactly $k$ times, then the remaining $n-k$ values can be determined in $c(n-k, p-rn, s-r)$ ways.
Suppose $n$ rolls satisfy the property that $n-m$ largest among them add up to $p$. Let $r$ be the smallest among the largest $n-m$ values. Let there be exactly $k$ rolls with value $r$ among the largest $n-m$. And let the number of rolls among the smallest $m$ rolls, whose value is equal to $r$ be $m-i$. Then the number of ways we can get such $n$ rolls is given by 
$$
\binom{n}{i} \binom{n-i}{k+m-i} (r-1)^i c(n-m-k,p-r(n-m),s-r)
$$
(The positions for $i$ rolls (whose values are less than $r$) and $k+m-i$ rolls (whose values are equal to $r$) can be picked in $\binom{n}{i} \binom{n-i}{k+m-i}$ ways. Each of these $i$ rolls can take any value less than $r$, so there are $(r-1)^i$ possibilities for that. And the remaining $n-m-k$ should all be greater than $r$ and add up to $p-rk$, hence the $c(.)$ term.)
The required answer is obtained by summing the above expression over all possible $i$, $r$ and $k$ values.
$$
\sum_{r=1}^{s} \sum_{k=1}^{n-m} \left[ c(n-m-k,p-r(n-m),s-r) \sum_{i=0}^{m} \binom{n}{i} \binom{n-i}{k+m-i} (r-1)^i \right]
$$
