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. 

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 $\epsilon \in \mathbb{N}, n, t, c \in \mathbb{N_0}$.)
