Number of permutations that can be made where the elements are in an increasing (not strictly) order up to a certain point I just found this question What is the number of permutations for given N numbers, such that the first part is non-decreasing? and it is phrased really badly, however it is actually quite interesting. It was solved by Frabala (I noticed the constraint right at the bottom of the OP's question that the largest value occurs at the index, I want to remove this constraint) I will also try to phrase the question better because I was quite lost.
First consider a set of numbers $\{1,...,n\}$, obviously there is only one way to arrange them in increasing size, this is where what he calls "the index" comes from. He wants to know how many permutations there where the first $i$ elements are increasing. He seems to do it backwards, by being given an $i$ and working out how many there are from that for small sets.
The set $\{1,2,4,5,3,...\}$ has "index" 4, because the first 4 are increasing. 
He allows duplicates, so the set $\{1,1,2,3,...,n\}$ is allowed, and in this case $\{1,1,2,3,5,4,...\}$ is an example of something of "index" 5.
There is an answer that comes close
In the answer by Frabala he has the following constraint:
 the i-th digit is the first occurrence of the largest number
 available in s

In this question at least I don't want this constraint, which is to say {1,2,3,4,5} has {1,3,2,4,5} as an example of "index" 2, where as with this constraint 5 would have have to be in the second position for all permutations of index 2. 
My question is:
Given a set of numbers how many ways are there of arranging them such that the elements up to a given index in the set are increasing (that is $a_n\le a_{n+1}$)
Given any set of numbers we can obtain a set of their positions in an ordering, that is {1,10,10,25} is the same as {1,2,2,3}, so the gaps don't matter. The OPs examples use gaps so I want to make it clear that they don't matter.
I am not wise enough to get Fabala's answer without help, the same goes for modifying it to remove the constraint, I've been playing about on paper and this is an interesting combinatorics question. 
 A: Let me rephrase the question so there is no ambiguity. Suppose we have a multiset $S$ of the numbers $1,2,\ldots, m$ each with multiplicity $a_i > 0$. Note that $\left|\,S\,\right| = n$ requires $\sum a_i = n$ and $m \leq n$. We say a permutation $\pi$ has index $q$ if its first descent is from $\pi_q$ to $\pi_{q+1}$, that is $\pi_1 \leq \pi_2 \leq \ldots \leq \pi_q > \pi_{q+1}$. We also assume that two elements of the same number are indistinguishable. This means that if $S = \{1,1,2\}$, then there are only three distinct permutations $\{112, 121, 211\}$. Our question is, how many permutations of $S$ have index $q$?
Let's first solve the non-repeating case $S = [n] =\{1,2,\ldots,n\}$. First we choose any $q\!+\!1$ elements to make up the prefix of our permutation $\{\pi_1, \pi_2, \ldots, \pi_{q+1} \}$. All but the largest of these elements can go in the $\pi_{q+1}$ spot, and the rest must be assigned in increasing order. So there are exactly $q {n \choose q+1}$ ways to fill the prefix. The suffix can then be any of the $(n\!-\!q\!-\!1)!$ permutations of the remaining elements. So the number of permutations of $[n]$ of index $q$ is
$$ q \; {n \choose q+1} \; (n-q-1)!$$
Now for the multiset case, there isn't going to be a clean solution since your answer depends on the multiplicities. Let $P$ be our prefix as before and let $\ell$ be the largest element in $P$. We can denote $P$ by the vector $(x_1, x_2, \ldots, x_{\ell})$ where $x_i$ specifies the number of occurrences of $i$. We note that $x_\ell \in [1,a_\ell]$ and the rest $x_i \in [0,a_i]$. We also must have $x_1 + x_2 + \ldots + x_{\ell} = q+1$. As before, $\pi_{q+1}$ can be chosen in $(q+1-x_\ell)$ ways and the rest of the prefix is determined. Letting $y_i = a_i - x_i$, the suffix is just any of the ${n-q-1 \choose y_1 \, y_2 \ldots \, y_m}$ permutations of the remaining elements. Summing over all possible prefixes
$$ \sum_{P} (q+1-x_\ell) {n-q-1 \choose y_1 \, y_2 \ldots \, y_m} $$
Unfortunately, to evaluate this sum you need to enumerate all the prefixes. I don't know if this can be simplified.
A: To answer "Given a set of numbers how many ways are there of arranging them such that the elements up to a given index in the set are increasing"
we can't simply  do this $\binom{n}{i}$, we will have to take in account the repetitions,
eg. if element 1 is repeated $r_1$ times, element 2 is repeated $r_2$ times, and so on, number of ways of choosing i-combinations will be given by
coefficient of $x^i$ in the expansion of
$(1+x + x^2+\cdots+x^{r1})\cdot(1+x + x^2+\cdots+x^{r2})\cdots (1+x + x^2+\cdots+x^{rn})$
If suppose out of 100, 60 elements are unique and rest 40 are being repeated r1, r2...times, then the number of ways become
coefficient of $x^i$ in the expansion of
$(1+x)^{60}\cdot (1+x + x^2+\cdots+x^{r1})(1+x + x^2+\cdots+x^{r2})\cdots (1+x + x^2+\cdots+x^{r40})$
There is no direct way to calculate this (if they are being repeated), only algorithmic approach will be able to find the coefficient of $x^i$ in this.
A: I just saw this question! :D
My answer at the question you're linking to is quite explanatory. I can't make it more explicit. Maybe if you play with some examples and try to follow the answer, it will get clearer to you. Or I hope someone here gives a better answer for you.
So, personally I will confine my answer to the modification needed to get a solution for your question. The only thing you need to change is that $S$ does not necessarily contain $l$. Therefore, you don't need to first select the first occurrence of $l$ and then $i-1$ numbers from the sequence produced from $s$ with the first occurrence of $l$ removed. You can just select $i$ numbers from $s$, with no restrictions.
$$\sum_{S\subseteq [1,100]~\text{and}~ |S|=i}\left[\binom{n}{i}\cdot\left(\prod_{a\in S} t^a_1!\right)\cdot\left(\prod_{a\in S} t^a_2!\right)\cdot\left(\prod_{a\notin S} t^a!\right)\right]$$
I have assumed that the numbers range from 1 to 100.
I would like to clarify that because a permutation is non-descending up to the $i$-th position, this doesn't mean that $a_{i+1}\leq a_i$ ($a_i$ is the number of the sequence in the $i$-th position). The non-descending ordering might continue after $i$, too, since the arrangement of the rest numbers is arbitrary. But, what we are looking for is the first part of the permutation until $i$, so we don't care if the non-descending order continues until some $i+k$.
