In Bollobás's Modern Graph Theory, section IV.4, Page 120 line -10, he uses the inequality
$$\frac{t {n \choose t}}{n {\epsilon n \choose t}} \leq \frac{t}{n} \epsilon^{-t} (1-\frac{t}{\epsilon n})^{-t}.$$
The general situation here is $t \approx \epsilon \ln n,$ if that's helpful. How do you derive this inequality?
To try to figure out this inequality, I applied Stirling's formula to the factorials in the lefthand side above. After some simplification (which I hope was correct!) I got
$$\frac{t {n \choose t}}{n {\epsilon n \choose t}} \sim \frac{t}{n} \epsilon^{-t} (1-\frac{t}{\epsilon n})^{-t} \left((1+\frac{t}{n-t})^{n-t+\frac{1}{2}} (1-\frac{t}{\epsilon n})^{\epsilon n + \frac{1}{2}} \right).$$
The rightmost parenthetical expression above seems to be larger than 1 (it takes the form big^big * small^small…). I'm not sure how to derive the inequality (either finishing this approach or otherwise).
Best Answer
Using the inequality $$ \frac{n-i}{\epsilon n - i} < \frac{n}{\epsilon n - t}$$ for $ 0 \leq i \leq t-1$, we have that $$\frac{t {n \choose t}}{n {\epsilon n \choose t}} = \frac{t}{n} \prod_{i=0}^{t-1} \frac{n-i}{\epsilon n - i} \leq \frac{t}{n} (\frac{n}{\epsilon n - t})^t = \frac{t}{n} \epsilon^{-t} (1-\frac{t}{\epsilon n})^{-t} $$.