Variance of number of cycles of length $t$ in a permutation We consider a uniform Distribution over all $n!$ permutations of $\{1, \dotsc, n\}$. 
Now we are interested in the Variance of the number $C$ of cycles of length $t$.
We have $$E[C]={n \choose t} \frac{t!(n-t)!}{t n!}=\frac{n! t! (n-t)! }{t n! (n-t)! t!}=\frac{1}{t},$$ since the probability of a fixed set building a cycle is $\frac{\frac{t!}{t}(n-t)!}{n!}$. 
Now, I wanted to compute the variance, but I am not sure whether it is correct.
We use $C=\sum_{S \subset \{1, \dotsc, n\}, |S|=t} X_S$, where $X_S$ is an indicator for the Event that $S$ builds a cycle.
$$Var[C]=E[C^2]-E[C]^2 = E[C]-E[C]^2 + \sum_{S \neq T} Pr[X_S=1 \wedge X_T=1]$$
$$=E[C]-E[C]^2+{n \choose t} {{n-t} \choose t} \frac{\frac{t!}{t}(n-t)!}{n!}\frac{\frac{t!}{t}(n-t)!}{n!}$$
$$=\frac{1}{t} - \frac{1}{t^2}+ {n \choose t} {{n-t} \choose t} \frac{\frac{t!}{t}(n-t)!}{n!}=\frac{1}{t}-\frac{1}{t^2}+\frac{n!(n-t)! t! (n-t)!}{t t! (n-t)! t! (n-2t)! n!}$$
$$=\frac{1}{t}-\frac{1}{t^2}+\frac{(n-t)!}{(n-2t)! t! t}=\frac{1}{t}-\frac{1}{t^2}+ \frac{{{n-t} \choose t}}{t}= \frac{1}{t}\left(1- \frac{1}{t} + {{n-t} \choose t}\right)$$
since only the disjoint $S$ and $T$ have non-Zero probability of both being a cycle. 
Is this correct now?
 A: Following the Wikipedia article about random permutations statistics, we may consider:
$$ \exp\left(\frac{x}{1}+\frac{x^2}{2}+\ldots+u\frac{x^t}{t}+\ldots\right)=\frac{1}{1-x}e^{(u-1)\frac{x^t}{t}} \tag{1}$$
hence the number of permutations with exactly $k$ cycles of length $t$ is given by:
$$ N(n,k,t) = n!\cdot [x^n]\frac{1}{2\pi}\int_{0}^{2\pi}\frac{1}{1-x}e^{(e^{i\theta}-1)\frac{x^t}{t}}e^{-ki\theta}\,d\theta$$
or, since:
$$\frac{1}{1-x}e^{(u-1)\frac{x^t}{t}}=\frac{1}{1-x}\sum_{j=0}^{+\infty}\frac{(u-1)^j x^{tj}}{t^j\,j!}$$
by:
$$ N(n,k,t) = n!\cdot [x^n]\left(\frac{1}{1-x}\cdot\sum_{j=0}^{+\infty}\frac{\binom{j}{k}(-1)^{j-k}x^{tj}}{t^j\,j!}\right)$$
that is:

$$ N(n,k,t) = n!\cdot\sum_{j=0}^{\left\lfloor\frac{n}{t}\right\rfloor}\frac{\binom{j}{k}(-1)^{j-k}}{t^j\cdot j!}=n!\cdot\frac{(-1)^k}{k!}\sum_{j=k}^{\left\lfloor\frac{n}{t}\right\rfloor}\frac{(-1/t)^{j}}{(j-k)!}=\frac{n!}{t^k\,k!}\sum_{j=0}^{\left\lfloor\frac{n}{t}\right\rfloor-k}\frac{(-1/t)^j}{j!}.\tag{2}$$

By $(1)$, we have that the expected number of cycles of length $t$ in a random permutation of $S_n$ is given by:
$$[x^n]\frac{\partial}{\partial u}\left.\frac{1}{1-x}e^{(u-1)\frac{x^t}{t}}\right|_{u=1}=[x^n]\frac{x^t}{t(1-x)}=\frac{1}{t}\cdot\mathbb{1}_{n\geq t}$$
and the variance can be computed through $(2)$ by approximating $N(n,k,t)$ with:
$$ N(n,k,t)\approx \frac{n!}{t^k\,k!}e^{-1/t}.$$
Hence, we are dealing with an approximated Poisson distribution having parameter $\lambda=\frac{1}{t}$, so the variance is expected to be $\frac{1}{t}$.
A: The answer is quite simple: if $2t\le n$ then the variance of the number of $t$-cycles of a random permutation is just $1/t$ (the same as the expectation), but  if $2t>n$ then the variance is $1/t-1/t^2$.
According to the original question, let $C(\sigma)$ be the number of $t$-cycles of any permutation $\sigma$ of $\{1,2,\dots,n\}$. Of course, we talk about expectation and variance because we understand the permutations as randomly taken, each one with uniform probability $1/n!$.
For any $t$-cycle $\tau$, write $\delta_{\tau,\sigma}=1$ if $\tau$ is a cycle in $\sigma$, $0$ otherwise. First, the computation of $\sum_\sigma C(\sigma)$ is
$$
\sum_\sigma C(\sigma) = \sum_\sigma\sum_\tau \delta_{\tau,\sigma} = \sum_\tau\sum_\sigma \delta_{\tau,\sigma} = \sum_\tau (n-t)! = (n-t)!\sum_\tau 1\,,
$$
for the number of permutations with $\tau$ as a cycle is just the number of permutations of the $n-t$ remaining elements. Now $\sum_\tau 1$, that is the number of $t$-cycles, is clearly ${n\choose t} \, (t-1)!$, and then
$$
\sum_\sigma C(\sigma) = (n-t)! (t-1)! {n\choose t} = \frac {n!} t\,,
$$
so (as said in the question) ${\rm E}(C) = \frac{\sum_\sigma C(\sigma)}{n!} = \frac 1 t\,.$
Next we  compute the variance: Recall that ${\rm Var}(C) = {\rm E}(C^2) - ({\rm E}(C))^2$.
Now
$$
\sum_\sigma (C(\sigma))^2 = \sum_\sigma\big(\sum_\tau \delta_{\tau,\sigma}\big)^2 = \sum_\sigma \big(\sum_\tau \delta_{\tau,\sigma}\big)\big(\sum_{\tau'} \delta_{{\tau'},\sigma}\big)\,.
$$
Note that $\delta_{\tau,\sigma}\delta_{\tau,\sigma}=\delta_{\tau,\sigma}$  so
$$
\sum_\sigma (C(\sigma))^2 = \sum_\sigma\sum_\tau \delta_{\tau,\sigma} + \sum_\sigma \big(\sum_{\tau\ne\tau'} \delta_{\tau,\sigma} \delta_{{\tau'},\sigma}\big) = \frac {n!} t + \sum_{\tau\ne\tau'} \big(\sum_\sigma \delta_{\tau,\sigma} \delta_{{\tau'},\sigma}\big)\,. 
$$
Two different cycles can appear in the same permutation only if they are mutually disjoint, and that permutation is determined by how we permute the remaining terms. Therefore:
(i) If $n<2t$ then $\delta_{\tau,\sigma} \delta_{{\tau'},\sigma}=0$ whenever $\tau\ne \tau'$, and we get that ${\rm E}(C^2) ={\rm E}(C) = 1/t$, so
$$
{\rm Var}(C) = \frac 1 t - \frac 1 {t^2}\,.
$$
(ii) In the case $n\ge 2t$, the last sum is
$$
\frac {n!} t + (n-2t)!\sum_{\tau\ne\tau'} 1= \frac {n!} t + (n-2t)!{n\choose t}{n-t\choose t}{(t-1)!}^2=n!\Big(\frac 1 t + \frac 1 {t^2}\Big)\,,
$$
so
$$
{\rm E}(C^2) = \frac 1 t + \frac 1 {t^2} \quad {\rm and} \quad {\rm Var}(C) = \frac 1 t\,.
$$
