Hat Puzzle - Probability, Expected Value, and Variance

[Aria1]

There are n sets of twins (so totally 2n people) attending a party. Each twin wears an identical hat to his/her twin sibling, and there are n different kinds of hat. Each person hands his/her hat to a hat clerk. When the party ends, unfortunately the hat clerk returns the hats to their owners completely at random. Let X be the random variable denoting the number of people who have their own hats back (since twins' hats are identical, getting either hat returned to them counts as getting their own hat back)

a) Find Pr{X=2n}
I think...
Pr{X=2n} = (2^n)/(2n)!
I am not entirely sure how to justify this/how this is achieved. I did it simply by setting n to be equal to different values and noticing the pattern. Is there some other way to look at it? If you could show me a mathematical process of achieving this result (or a different one if this is wrong), that would be wonderful!

b) Find E(X)
I think...
E(X_i) = 1/n
so
E(X) = 2n(1/n)
= 2

c) Find Var(X)
I think...
Var(X) = SUM_{i=1 to i=2n}(Var(X_i)) + DOUBLESUM_{i=1 to i=2n and j=1 to j=2n where i not = j}(Cov(X_i,_j). Then Var(X_i) = (1/n) - (1/n²), and SUM = 2n((1/n)-(1/n²). I am not positive about this though, and do not know how I could go about finding the Cov(Xi,Xj) and summing it - how many summands of covariance does the double-sum term consist of?

Any help particularly with Part C is greatly appreciated - thanks!