$n \geq 4$ men, among whom are A, B and C stand in a row.then what is the probability that C stands somewhere between A and B $n \geq 4$ men, among whom are A, B and C stand in a row. Assume that all possible orderings of the $n$ men are equally likely, then what is the probability that C stands somewhere (not necessarily adjacent to A and B) between A and B.
Well, I know the number of orderings is $n!$ but here it seems we don't need to use this. Since they are placed equally, it means the probability of A between B and C, and the probability of B between A and C, and the probability of C between A and B should be equal. So the answer should be $\frac{1}{3}$?
Thanks.
 A: Your explanation is correct. To make it a bit more explicit, consider the positions 3 occupied by $A$, $B$, and $C$, regardless of how the three are arranged within those positions. there are $\binom n3$ such positions, and for each there are $6$ orderings of $A$, $B$, and $C$ from leftmost to rightmost. By symmetry, each is equally likely.
$$ABC,ACB,BAC,BCA,CAB,CBA$$
Thus in each of the possible placements of the three, two of the six possible orderings place $C$ in the center. The probability is thus $\frac 13$.
To think about this another way, you can place $A$, $B$, and $C$ first, at which point the pobability that $C$ is in the center is clearly $\frac 13$. Then insert the other $n-3$ individuals into the line without any swapping of positions. The ordering, and the probability of $C$ being in between $A$ and $B$, will not change in this process.
A: Comment:  For doubters (if any), a simulation with $n = 10$. Identify $A, B,$ and $C$
with $1, 2,$ and $3,$ respectively. Perform the experiment a million times.
Orders "$A$ left of $B$ left of $C$" and "$C$ left of $B$ left of $A$" each 
have simulated probabilities very near 1/6, as claimed. [(+1) for Answer of 
@Kajelad.]
m = 10^6;  n = 10;  men=1:n
ord.1 = ord.2 = ord.3 = numeric(m)
for (i in 1:m) {
 perm = sample(men, n)
  ord.1[i] = which(perm==1)   # position of '1' in line
  ord.2[i] = which(perm==2)   # etc.
  ord.3[i] = which(perm==3)  }
mean(ord.1 < ord.2)
## 0.50057                    # aprx P(A left of B) = 1/2
mean(ord.1 < ord.2 & ord.2 < ord.3)
## 0.167329                   # aprx P(Order A < B < C) = 1/6
mean(ord.3 < ord.2 & ord.2 < ord.1)
## 0.166655                   # aprx P(Order C < B < A) = 1/6

