How many final placements are possible if the leader changed 0...5 times? Six competitors A-F take part in cycling time trial. The race is about covering several dozen kilometers as quickly as possible. Following the rules of this sporting event, competitors start one by one, at regular intervals. Let's call the cyclist, who has ridden the entire route and at the moment he is in the lead (so he had the best time so far), the leader. The final placement is the list obtained by ranking the cyclists according to the final places they took in the race. How many final placements are there in which:

*

*A) the leader didn't change from the start;

*B) the leader changed 2
times;

*C) the leader changed 3 time;s

*D) the leader changed 4
times;

*E)the leader changed 5 times;

I believe the key is understanding the mechanic of leadership. Also they don't wait for the previous one to finish so for example B can finish before A. Then A can no longer take the leadership. Subproblems A and E are trivial, A is just 5! and E is 1, if I am not mistaken.
EDIt: I got more info, answers for 3 changes and 4 changes are 85 and 15 respectively. Looking for a way that isn't bruteforcing it.
 A: You are correct for A.  The first rider has to stay in front, but the others can be in any order.
For B, any of the riders except A can be the eventual winner.  If B is the eventual winner, the others can finish in any order, so there are $5!$.  If C is the eventual winner, B cannot beat A, so there are $\frac 12 \cdot 5!$.  If D is the eventual winner, B and C cannot beat A, so there are $\frac 13 \cdot 5!$.  If E is the eventual winner, $\frac 14\cdot 5!$ and if F is the eventual winner, $\frac 15 \cdot 5!$ for a total of $\frac {137}{60}\cdot 120=274$.  $\frac {137}{60}=H_5$, the fifth harmonic number.
For C you need to look at the two riders that take the lead.  As in B, riders before the first to take the lead cannot beat A.  Now riders between the first to take the lead and the eventual winner can beat A but not the interim leader.
A: Label the riders $1,2,\ldots,n$ according to their start order. Their final placings determine a permutation $\pi=r_1r_2\ldots r_n$ of $[n]$, where $r_1$ is the winner, $r_2$ has the second-best time, and so on. For $k\in[n]$ say that $r_k$ is a left-to-right minimum if $r_i>r_k$ for $i<k$. If $r_1r_2\ldots r_n$ has $m$ left-to-right minima, the lead changed hands $m-1$ times.

For example, the left-to-right minima of the permutation $463512$ of $[6]$ are $4$, $3$, and $1$. The lead changed hands first from $1$ to $3$ and then from $3$ to $4$.

If the left-to-right minima are $r_{k_1}=r_1,r_{k_2},\ldots,r_{k_m}$, insert $m$ pairs of parentheses to get
$$(r_{k_1}\ldots r_{k_2-1})(r_{k_2}\ldots r_{{k_3}-1})\ldots(r_{k_m}\ldots r_n)\tag{1}$$
and interpret $(1)$ as a permutation $\varphi(\pi)$ in cycle notation. It’s well-known and easy to see that the map $\varphi:S_n\to S_n$ is a bijection: given a permutation $\pi\in S_n$ in cycle notation, rotate each cycle to bring the smallest element to the front, order the cycles in descending order of smallest elements, and remove the parentheses to get $\varphi^{-1}(\pi)$. Thus, the number of outcomes in which the lead changed hands $k$ times is the number of permutations of $[n]$ with $k+1$ cycles.
This is given by $n\brack{k+1}$, a Stirling number of the first kind. These do not have a nice closed form, but they do satisfy a nice recurrence:
$${{n+1}\brack k}=n{n\brack k}+{n\brack{k-1}}$$
for $k>0$, with initial values ${0\brack 0}=1$ and ${n\brack 0}={0\brack n}=0$ for $n>0$.
For the present case with $n=6$ we have
$${6\brack 1}=120,{6\brack 2}=274,{6\brack 3}=225,{6\brack 4}=85,{6\brack 5}=15,{6\brack 6}=1\,.$$
To sum up, $n\brack k$ is the number of possible outcomes when there are $n$ riders, $k$ of whom are leaders at some point during the time trial; this is the number of possible outcomes for $n$ riders when the lead changes hands $k-1$ times.
