Consider a balls and bins experiment with $2n$ balls but only two bins. Each ball is thrown independently into a bin chosen uniformly at random. Let $X_1$ be the random variable for the number of balls in bin $1$ and $X_2$ for bin $2$. It is easy to see that $E[X_1] = E[X_2] = n$. We would like to have a handle on the difference $X_1 - X_2$. Our goal is to prove that for any fixed $\epsilon > 0$ there is a fixed constant $c > 0$ such that $\Pr[X_1 - X_2 > c \sqrt{n}] < \epsilon$. By symmetry we can then argue that $\Pr[|X_1 - X_2| > c \sqrt{n}] < 2 \epsilon$.
1.Compute the variance of $X_1$. Then use Chebyshev bound to show that $\Pr[|X_1 - n| > c \sqrt{n}] \le \epsilon$ for suitable choice of $c$ for a given $\epsilon$. What is the dependence of $c$ on $\epsilon$?
2.Use Chernoff bound to show that $\Pr[|X_1 - n| > c \sqrt{n}] \le \epsilon$. You need to use the bound separately for computing $\Pr[X_1 > n + c \sqrt{n}]$ and for $\Pr[X_1 < n - c \sqrt{n}]$. What is the dependence of $c$ on $\epsilon$?
3.Using the preceding show that $\Pr[X_1 - X_2 > c \sqrt{n}] < \epsilon$.
So, I have already compute the variance of $X_{1}$, and shown part (1) and part (2), but not sure how to use them to show (3)