Let $X\_{i} = 1$ if *ith* person gets his or her own hat back and $0$ otherwise. Let $S\_{n} = \sum\_{i=1}^{n}X\_{i}$, then $S\_{n}$ is the total number of people who get their own hats back. Show that:

a) $E(X\_{i}^{2}) = \frac{1}{n}$

b) $E(X\_{i}.X\_{j}) = \frac{1}{n(n - 1)}$ for $i\neq j$

c) $E(S\_{n}^{2}) = 2$ (using (a) and (b)).

d) $var(S\_{n}) = 1$