There are $2n$ people at a party, making $n$ couples. These $2n$ people are seated randomly at $n$ tables that have $2$ seats each. Let $X$ be the number of couples that sit at the same table. Find $E[X]$ and $var(X)$.
I tried to use $$P(X=n-2k)=C(n,n-2k)n!(2^n)/(2n)!$$ for $k\in[0,n/2]\in\mathbb{N}$ and $0$, otherwise
so that $$E[X]=\sum P(X=n-2k)(n-2k)$$ but I feel like I'm missing something. Is there a better way to calculate $f_X(x)$?
It's easier to just calculate the expectation and the variance instead of finding the whole distribution.
For the expected value, try using linearity of expectation, applied to indicator random variables for the events "couple $i$ is sitting together".
For the variance, one approach is to use the fact that $$\text{Var}(\displaystyle\sum_{i = 1}^n X_i) = \sum_{i,j}^n \text{Cov}(X_i,X_j)$$ where again each $X_i$ is an indicator random variable for the event "couple $i$ is sitting together". When $i = j$, $\text{Cov}(X_i,X_j) = \text{Var}(X_i)$ and is easily evaluated. When $i \neq j$, $\text{Cov}(X_i,X_j) = \mathbb{E}[X_iX_j] - \mathbb{E}[X_i]\mathbb{E}[X_j]$. These are probabilities that are not too hard to calculate.