Divisible binomial coefficients

[noowutah]

Homework Statement

I need to sum the binomial coefficients that are divisible by a
positive integer t, i.e.

$$\sum_{i=0}^{s}\binom{ts}{ti}$$

Is there any way to get rid of the sum sign?

Homework Equations

Let t be fixed and s go to (positive) infinity (both t and s are
positive integers). Let M(s) be a set with #M(s)=ts, then I am really
interested in the expected value of the number of elements when you
choose subsets from M whose cardinality is a multiple of t. For
example, what is the mean number of elements picking subsets with
cardinality 0, 3, 6, or 9 from a set with cardinality 9 (t=3, s=3)?
Where does this expected value go as s (the ``grain'' of M) goes to
infinity?

$$EX=\frac{\sum_{i=0}^{s}ti\binom{ts}{ti}}{\sum_{i=0}^{s}\binom{ts}{ti}}$$

The Attempt at a Solution

I anticipate the solution to be lim(s->infty)EX(s)=ts/2, but I'd love
to prove it.

 

[noowutah]

binomial coefficients problem

Clarifying and rephrasing:

I have two finite sets A_{1} and A_{2} with the same number of
elements (let their cardinality be s times t, where t is a fixed
positive integer). Let me randomly pick elements from these two sets,
with one constraint, however: the number of elements picked from A_{2}
must be a t-multiple of the number of elements picked from A_{1}. If
t=3, for example, and s=2, there are six elements in A_{i} and I can
either pick 0 elements from A_{1} and 0 from A_{2} (there is only one
way of doing this), or 1 element from A_{1} and 3 elements from A_{2}
(there are 120 ways of doing this), or 2 element from A_{1} and 6
elements from A_{2} (there are 15 ways of doing this). Let X be the
random variable counting the elements picked from both sets. In
the example, X can be 0, 4, or 8, and the associated probabilities are
1/136, 120/136, and 15/136, so that the expectation for X is EX=4.41.

I want to know what this expectation is for fixed t and variable s as
s increases. I can provide the formula for fixed s and t, but I have
no idea how to investigate the behaviour of this formula as s increases.

$$EX=(1+t)\frac{\sum_{i=0}^{s}i\binom{ts}{i}\binom{ts}{ti}}{\sum_{i=0}^{s}\binom{ts}{i}\binom{ts}{ti}}$$