Let $n>15$ be an integer. Suppose also $n=\sum\_{i=1}^n ic\_i$, where $c\_i$ are non-negative integers. Assume further that $c\_1<4$.


Is the following inequality true?
$$\frac{n!}{\prod\_{i=1}^{n}i^{c\_i}\prod\_{i=1}^{n}c\_i!}>n(n-1)(n-2)(n-3)$$.


Motivation: The left hand side of the inequality is the size of the conjugacy class of a permutation $\sigma$ with cycle decomposition $n=\sum\_{i=1}^n ic\_i$, i.e., $\sigma$ is the product of $c\_1$ cycles of length $1$, $c\_2$ cycles of length $2$,.... We have encountered the question in studying a group $G$ of order $|A\_n|=n!/2$ having the same multiset of conjugacy class sizes as the alternating group $A\_n$ of degree $n$. We are trying to show that any non-trivial normal subgroup of $G$ is actually the whole $G$, that is, $G$ is simple.
Why $n>15$, this is maybe more technical; but let me say that the problem for less than $n\leq 15$ hase been solved. We did not chech if the inequality valid or not for $n\leq 15$.



Factorial-inequalities
Note that for integers $i > 2$ and. $h > 0$, and except for the case $(i,h) = (3,1)$, one has 
$i^h(h!) < 2^{\lfloor ih/2 \rfloor}(\lfloor ih/2 \rfloor !)$. We can account for the exception and bound 
from above the denominator of the left hand side of the posted inequality by 
$(3)2^{(n - c\_1)/2}(((n - c\_1)/2)!)(c\_1!) \leq (3)2^{(n - c\_1)/2}(((n + c\_1)/2)!)$. This latter term is increasing
in $c\_1$, and for large $n$ one can have $c\_1 \leq n - 10$ and still be less than $(n - 4)!$. One can use
this to show the inequality of the post is satisfied for $n > 15$ and $c\_1 < 4$; likely the inequality holds 
for more $n$ and more $c\_1$.


**BEGIN EDIT 2012.01.05**
I decided to add some detail to the post.


Letting $f(i)=i^{c\_i}(c\_i)!$, we can rearrange the poster's inequality to $$(n - 4)!  \gt \prod\_{1 \le i \le n}f(i)$$, and ask for which values of $n, i,$ and $c\_i$ the inequality holds.  Given $c\_i$, let $g(i) = 2^{\lfloor ic\_i/2 \rfloor} (\lfloor ic\_i /2 \rfloor)! $ for $i \gt 1$ and $g(1) = f(1)$.  Now, when $i = 3$ and $c\_i = 1$ we have $f(3) \lt 2g(3)$, and for other pairs $(i,c\_i)$ with $i \gt 2$ and $c\_i \gt 0$ one has $f(i) \le g(i)$, so one can have the original inequality follow from $$(n-4)! \gt 2\prod\_{1 \le i \le n} g(i) $$.   However, the product of the $g(i)$ is itself bounded from above by $h=2(c\_1)! 2^{(n - c\_1)/2} {\lceil(n - c\_1)/2\rceil}!$.  When $c\_1 \le 4$ and $n \ge 15$,  $2h$ is less than $(n-4)!$.   


It is clear that the original inequality implies $c\_1 \lt n-4$, and that there is no solution for $n \lt 8$.  For $n=8$, the product of the $f(i)$ has to be less than $24$, so $c\_1 \lt 4$. For $n= 8,9,10,11$ it is routine to find restrictions on $c\_1$ that will permit solutions of the inequality. From another direction,  if $c\_1 = n-5$ then case-by-case examination gives that $f(i) = 1$ for $ i \gt 5$ and $f(2)f(3)f(4)f(5)$ is at most $8$, so that $n > 12$ for the original inequality to hold.  When $c\_1 = n - 6$, a similar analysis requires $n > 11$, and for larger values of $n - c\_1$ the inequality holds for all the remaining meaningful cases. 
**END EDIT 2012.01.05** 


Gerhard "Ask Me About System Design" Paseman, 2011.12.26


