How does this amount change if we add to it? We can iterate over the naturals, with zero included.  Here the focus is on the numbers from $0$ to $2^s - 1$, inclusive, in binary.  So we have the numbers as:
$$0000, 0001, 0010, \dots 1111 \text{ for } s=4$$
We then have an ordered set of urns of size $s$.  We can pick $a$ of the urns, and label them with a $1$, and label the rest as don't care, with an $x$.  We can again iterate through all possible ordered combinations of urns.  So, for $s=4, a=2$, we have:
$$11xx, 1x1x, 1xx1, x11x, x1x1, xx11$$
as labels for the urns.
We then try to match up each number, in binary, with the ordered set of urns.  So, for example, $1101$ would fit in the urn labeled $11xx$ since the ones match, and the $x$s don't matter.  However, $0101$ would not fit in the urn $11xx$, since there is a zero in the leftmost digit, but the urn has a one in the leftmost digit.
Now, if we count the matches between all combinations of numbers and urns, there is a formula that counts how many matches there are that have $m$ ones in the digits.  I'm including this as a check to ensure that I've described the setup properly.  It is:
$$\binom{s}{a}\cdot\binom{s-a}{m-a}$$
THE QUESTION
Now, we suppose that we enlarge the numbers by attaching $c$ ones to them.  If we iterate over all possible combinations of picking $s$ digits from the newly enlarged numbers, what is the new formula for the number of valid matches?
AN EXAMPLE
If $s=5$, we get numbers that are the binary values of the numbers from $0$ to $2^5-1$, inclusive.  So the numbers that we have are:
$$00000, 00001, 00010, \dots, 11111$$
The numbers all have 5 binary digits, so there are 5 urns.  Now we can pick a number $a$, which is the amount of urns that will be labeled with a $1$.  So again, if $a$ is two, we have urns labeled as:
$$11xxx, 1x1xx, 1xx1x, 1xxx1, x11xx, x1x1x, x1xx1, xx11x, xx1x1, xxx11$$
Then we are trying to count the amount of numbers that will fit in the urns.  For example, the number $00000$ won't fit into any of the urns, since it simply doesn't have enough ones as digits.  On the other hand $11100$ will fit into $11xxx$, since it has ones in the two leftmost positions.  $11100$ will also fit into $1x1xx$ since it has two matching ones in the correct positions.  $11100$ will also fit into $x11xx$, since it again has two ones in the correct positions.
For the formula so far, this would give $\binom{s}{a}$ ways of choosing the ones and $x$s, and then $\binom{s-a}{m-a}$ ways of choosing the numbers that fit.
The New Problem
We are adding a variable $c$ to the equations.  We can suppose that $c=3$.
Now we basically add three ones to the left side of every number.  So for example, if the original number is $9$, which is $01001$ in binary, we add $c=3$ ones to the left side to get $11101001$.  We also expanded the size of the urns to accommodate for the new numbers.  So if $a=2$, the new urns are labeled as:
$$11xxxxxx, 1x1xxxxx, 1xx1xxxx, 1xxx1xxx, \dots$$
...Essentially, we go through every possible combination of labeling $a=2$ urns with a one, and the rest as don't cares.
Now we want to get a new formula for these expanded values.
SOME NOTES
I believe the new formula will be somewhat similar to the old.  The main difference is that we increase the size of the numbers (and urns) by $c$, while at the same time, forcing the $c$ leftmost digits to be $1$.
As for the number $a$, it represents the amount of urns that are labeled with a one.  It is in both formulas.  It simply tells us how many of the urns, in both equations, that we must label with a one.  As another example, if there are 5 urns, and $a=4$, the possible combinations of urns are:
$$1111x, 111x1, 11x11, 1x111, x1111$$
 A: In the new problem, when we choose a configuration with a 1's and the rest x's, it is important to know whether the 1's occur in the c-block to the left of the number or in the rest. The reason is, suppose we start with s = 3 digits and c = 2, and also a = 2.
Then the string xx11x only matches 11110 and 11111, (one string with 4 1s and one string with 5). since in getting the new strings we were required to add 1's to the left. On the other hand, 11xxx matches eight strings: 11000, 11100, 11010, 11001, 11110, 11101, 11011, 11111. (I hope I'm understanding this problem correctly..)
So the formula will be written as a sum over how many of the a 1's occur in the left c-block in the urn label. Suppose this number is k. Then $a - k$ 1's appear in the right s-block. There are $\binom{c}{k}\binom{s}{a - k}$ ways to get our urn label. Now  we want to count the number of matching strings with m total ones, then there are $s - (a - k)$ undetermined slots (marked with $x$'s) and $(m - c) - (a - k)$ of them have to be 1's. The formula is therefore:
$$\sum_{k}\binom{c}{k}\binom{s}{a - k}\binom{s - (a - k)}{(m - c) - (a - k)}$$
The constraints on the variables are: $s + c \geq m \geq c$ (the number of 1's must be at least $c$ and at most the length of the string $s + c$), $m \geq a$, and $s \geq a$. So what's the range on $k$: Well $k$ can be at most $c$, but it can also be no greater than $a$, and so we can write the formula slightly better:
$$\sum_{k = 0}^{min(c,a)}\binom{c}{k}\binom{s}{a - k}\binom{s - (a - k)}{(m - c) - (a - k)}$$
