Divisible binomial coefficients

noowutah
Messages
56
Reaction score
3

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.
 
Physics news on Phys.org
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}}
 
Last edited by a moderator:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
4
Views
2K
Replies
2
Views
2K
Replies
9
Views
2K
Replies
5
Views
2K
Replies
2
Views
2K
Replies
1
Views
1K
Replies
43
Views
4K
Back
Top