There are n husbands and n wives to be seated at... Find var(X) There are n husbands and n wives to be seated at a long rectangular
table. All of the husbands will sit on one side of the
table and all of the wives will sit on the other side. Let X be
the number of husbands that sit across from their wives. Find
var(X). 

I've setted $X_i=1$ If husband seats across his wife and $X_i=0$ otherwise 
(with $i=1,2....n$).
Also, $P(X_i=1)$=$1\over n$, 
I then have: $E(X_i)=\sum_{k=1}^{n}x_iP_X(x_i=1)=1$.
From the variance definition, I can have: $Var(X_i)=E(X_i^2)-(E(X_i))^2$
After that, I'm a bit stuck. 
Can I just continue with my last line

$Var(X_i)=E(X_i^2)-(E(X_i))^2$

and then apply a $\sum_i^n$ on the expectations? 
Or can I just write more on the line of: 

$Var(X_i)= \frac{n-1}{n^2}$ with $i\in [0,n]$

Any help or tip would be greatly appreciated.
 A: Comment. Because you can google several proofs online, including the one
in my earlier comment for the 'hat check' incarnation, maybe it is worthwhile
to illustrate the reasonably good fit to $\mathsf{Pois}(1)$ using simulation
in R. With a million iterations one can expect at least two place accuracy.
set.seed(412);  m = 10^6;  n = 12
x = replicate(m, sum(sample(n)==1:n))
mean(x);  sd(x);  table(x)/m
[1] 1.001044       # aprx E(X) = 1
[1] 0.999903       # aprx SD(X) = 1
x
       0        1        2        3        4        5        6        7        8 
0.367157 0.368324 0.184228 0.061268 0.015332 0.003110 0.000495 0.000076 0.000010 

hist(x, prob=T, br=(-1:n)+.5, col="skyblue2")
k = 0:n; pdf = dpois(k, 1);  points(k, pdf, col="red")

The histogram below shows the simulated distribution of $X$ and the
centers of red circles show probabilities of $\mathsf{Pois}(1).$

Note:  By one of @TheoreticalEconomist's links, $P(X = 0) = \sum_{i=0}^{12} (-1)^i/i! = 0.3679.$
i=0:12;  sum((-1)^i/factorial(i))
[1] 0.3678794

