Dear All,


I am considering maximal subgroups of odd index in Alternating and Symmetric groups, and this leeds me to some questions on binomial coefficients that I presently do not know and that I need some hints and suggestions :


1. Assume that $ab=2^t$ with $a>1, b>1,t\geq 3$. Which values of $a$ and $b$ such that
$$\frac{(2^t)!}{(a!)^b.b!}$$
is odd and smallest.
2. Let $n\geq 9$ be an odd integer (not prime) and $p$ the largest prime less than $n$. Is there always an integer $r\geq 1$ such that $\binom{n}{r}$ is odd and divisible by $p$. Note that this is not true for the case $n=9$. Is it also true for $n$ even?


Thanks you very much in advance for comments and advice.



primes dividing binomial coefficients
Hi! I have a proposed answer for 2). 


You can construct an infinite series of counter examples. [Erdos proved that there are primes between $n$ and $2n$](http://en.wikipedia.org/wiki/Bertrand%27s_postulate). Now consider $2n=2^k$ for some $k>2$. 


We can observe that the binomial number $\binom{2n+1}{r}, 1< r < n+1$ is always even. 
The upper product have powers of 2 as the following sequence: $\lbrace k,1,2,1,3,\dots,s \rbrace$  

while the bottom product have powers of 2 of the form: $\lbrace 1,2,1,3,\dots,s,k-1\rbrace$.  

This can be reflected for cases $\binom {2n+1}{r}, n < r < 2n$.


The difference of the sum of these 2 sequences is at least 1.
As a result, the binomial number may be divisible by $p$ but it will not be even. 


(Edit) Exception: When the binomial number is $\binom{2n+1}{1}$ or $\binom{2n+1}{2n}$ you have either no powers of 2 or same powers of 2. But in this case it will not be divisible by $p$.


