Is $n\binom{\epsilon n}{t}>t\binom{n}{t}$ for large $n$ and fixed $\epsilon$ and $t$ Let $\epsilon$ and $t$ be fixed numbers with $t$ and integer. I came across the following inequality in a counting problem. 
$$n\binom{\epsilon n}{t}>t\binom{n}{t}.$$
I want to show that for $n$ large enough this inequality is true. I don't know if my hypothesis is true or not. I've made many lists in mathematica with different $\epsilon$ and $t$. All of the lists showed my hypothesis was correct in that case. Any ideas on how to prove this result? Any much appreciated. 
 A: Intuitively, the left hand side grows linearly in $n$ and has a bigger binomial for any $n \in \mathbb{N}$ (since $\binom{n}{c} \geq \binom{m}{c}$ for any $c$ and $n\geq m$). Formally:
$$n\binom{\epsilon n}{t}= n \frac{(\epsilon n) ^{\underline t}}{t!}$$
and 
$$t\binom{n}{t}= t \frac{n ^{\underline t}}{t!}$$
we get
\begin{align}
n\binom{\epsilon n}{t} &> t\binom{n}{t} \\
\iff n \cdot (\epsilon n) ^{\underline t} &> t \cdot n ^{\underline t}
\end{align}
Due to the monotonicity of the falling factorial for any $t \geq 0$, $\epsilon n \geq n$ and $n > t$ for large $n$, your inequality follows.
(All this is assuming that $\epsilon \in \mathbb{N}, n, t, c \in \mathbb{N_0}$.)
A: Let $\epsilon>0$ and $n,t\in\mathbb{N}$. Assume for sake of contradiction that for all $n$
$$n\binom{\epsilon n}{t}\leq t\binom{n}{t}.$$
Then that would yield 
$$ \binom{\epsilon n}{t}/\binom{n}{t}\leq \dfrac{t}{n}.$$
for all $n$. Using the appropriate approximations from this wikipedia page 
 http://en.wikipedia.org/wiki/Binomial_coefficient
We have 
$$ \dfrac{\epsilon^{t}}{t^{t}t!}\leq \dfrac{t}{n}$$
for all $n$ which is clearly false because $\epsilon $ and $t$ are fixed.
