Number of permutations$\pi\in S_{30}$ with $\pi(2)<\pi(3)$

Question

I am trying to find the number of permutations $\pi\in S_{30}$ with $\pi(2)<\pi(3)$.

I worked out the answer as: $28!\times (1+2+\cdots+29)$.

My reasoning is: Suppose $\pi(2) = 1$ then there are $29$ possibilities for $\pi(3)$ and for each of these $29$ possibilities there are $28!$ permutations of the rest of the numbers $1,3,4,...,30$. We can do the same if $\pi(2) = 2$, but in this case the number of possibilities for $\pi(3)$ are $28$ and there are again $28!$ permutations for the rest of the numbers etc. Hence in total there are $28!\times (1+2+\cdots+29)$ possible permutations?

Answer 1

Your argument is correct. You can simplify your answer by writing $$1+2+\ldots+29=\frac{1}{2}\times29\times30.$$ Another way to get to the same answer, is by noting that either $\pi(2)<\pi(3)$ or $\pi(2)>\pi(3)$ holds for every permutation $\pi\in S_{30}$.

Answer 2

The map $$\iota:\quad {\cal S}_{30}\to{\cal S}_{30},\qquad \pi\mapsto \pi\circ\tau_{23}\ ,$$ where $\tau_{23}$ denotes the transposition of $2$ and $3$, maps the set $A$ of permutations satisfying $\pi(2)<\pi(3)$ onto its complement. It follows that $A$ has exactly ${1\over2}\bigl|{\cal S}_{30}\bigr|$ elements.

Answer 3

Let $A$ be the set of permutations $[x_1, x_2, \ldots, x_{30}]$ in $S_{30}$ such that $x_2 < x_3$ and let $B$ be the set of permutations $[x_1,x_2,\ldots,x_{30}]$ in $S_{30}$ such that $x_2 > x_3$. Note that $|A|=|B|$ because the map from $A$ to $B$ that takes $[x_1,x_2,x_3,\ldots,x_{30}]$ to $[x_1,x_3,x_2,\ldots,x_{30}]$ (ie it interchanges the second and third components) is a bijection. The set of all permutations in $S_{30}$ is a disjoint union of the two sets $A$ and $B$, whence $|A| = |S_{30}|/2$ = (30!)/2$.