Are these two events in $S_{30}$ independent? Let $S = S_{30}$, the set of permutations on $\{1,...,30\}$.
Suppose $A$ is the event: $\{\pi\in S_{30}: \pi(1) = 1\}$
and $B$ is the event $\{\pi\in S_{30}: \pi(2) < \pi(3)\}$.
I am asked to show that these events are independent.
Now $|A| = 29!$ and $|B| = (1+...+29)\times 28!$
Hence $|A|\times|B| = 29!\times 28! \times (1+...+29)$
However $|A\cap B| = (1+...+28)\times 27!$
So $P(A\cap B)< P(A)\times P(B)$.
So these events don't seem to be independent? What am I doing wrong?
 A: I assume that we are picking a permutation $\pi$ at random, with all permutations equally likely.
Let $A$ be the event $\pi(1)=1$ and $B$ the event $\pi(2)\lt \pi(3)$. To show independence, we show that $\Pr(B\mid A)=\Pr(B)$. This is clear, for by symmetry each is $\frac{1}{2}$.
A: By definition: since $\;A\cong S_{29}\;$ :
$$P(A)=\frac{29!}{30!}=\frac1{30}$$ , whereas
$$B:=\{\pi\in S_n\;/\;\pi(2)<\pi(3)\}$$ 
and we can then count:
$$\begin{cases}\pi(3)=30\implies\pi(2)=1,2,...,29\\{}{}\\\pi(3)=29\implies\pi(2)=1,2,...,28\\\ldots\\\pi(3)=2\implies \pi(2)=1\end{cases}$$
For each option in each line we have $\;(n-2)!\;$ permutations, so all in all we have
$$28!\left(29+28+\ldots2+1\right)=28!\frac{29\cdot30}2=\frac{30!}2\implies$$
$$P(B)=\frac{\frac{30!}2}{30!}=\frac12$$
Finally: if $\;\pi\in A\cap B\;$ then
$$\begin{cases}\pi(1)=1,\,\pi(2)=2\implies \pi(3)=2,3,....,30\\{}\\\pi(1)=1,\,\pi(2)=3\implies \pi(3)=4,...,30\\\ldots\\{}\\\pi(1)=1,\,\pi(2)=29\implies \pi(3)=30\end{cases}$$
For each option above we have $\;27!\;$ permutations, so in total we have
$$27!\left(1+2+\ldots+28\right)=\frac{29!}2\implies P(A\cap B)=\frac{\frac{29!}2}{30!}=\frac1{60}=P(A)P(B)$$
and the events are independent
