Expectation of the number of men seated between two specific women I’ve been having a hard time answering this question (I’m bad at combinatorics) - i’d like it if you could help!  The Question

$m \geq 1$ men and $n \geq 2$  women sit randomly on a bench with $m+n$ places, two of these women are Hilla and Nikki. What is the expectation of the number of men sitting between Hilla and Nikki?

So far my thinking was to look at particular cases and then move on to a generalized expression I can calculate. 
I think that for $i$ men between Hilla and Nikki I can look at the permutations as
$(n+m-i-2+1)! = (m+n-1-i)!$ “outside” permutations times “inner” permutation of the Hilla/Nikki area. $(m)(m-1)(m-2)\dots (m-i+1) = \frac{m!}{(m-i)!}$ for picking i men since order matters, and multiply it by $2$ since I can switch Hilla and Nikki around. So overall the expectation is $$E(M)=\sum_{i=1}^m i \cdot 2ֿֿ\cdot \frac{m!}{(m-i)!} \cdot (m+n-1-i) ! \cdot  \frac{1}{(n+m)!}$$
but I think I didn’t account for the possibilities of women sitting between them.
Overall this seems like a really complicated approach.
 A: For each man the probability of him sitting between Hilla and Nikki is exactly $\frac13$, since there are $6$ equally likely ways to permute them alone and $2$ of the ways have the desired result. Then by linearity of expectation the expected number of men between the two named women is $\frac m3$.
A: Parcly Taxel's approach is much more elegant than mine.  My approach:
You can assume that the $(n-2)$ women, other than Hilla and Nikka do not exist, for the purposes of computing the expected number of men sitting between Hilla and Nikka.  That is, regardless of how many women that there are other than Hilla and Nikka, and regardless of where they are sitting, Hilla and Nikka will (in effect) create $(3)$ regions.  These are the regions before and after Hilla and Nikka, and the region between them.  The seating of the $(n-2)$ other women has no effect on the distribution of the men among these $(3)$ regions.
One way of visualizing this is to analogize to boarding an airplane, which is done in ordered groups.  First, Hilla and Nikka are seated, then the men are randomly seated in the three regions, and then the $(n-2)$ women are seated in which ever seats are remaining.  So, the seating of the women, which (hypothetically) occurs after the seating of the men, does not affect the location of the men, with respect to Hilla and Nikka.
So, ignoring the $(n-2)$ other women, there are $(m+2)$ distinct ways that the men, and Hilla and Nikka, can be permuted.
For $r \in \{0,1,2,\cdots, m\}$, let $f(r)$ denote the number of ways that there can be exactly $(r)$ men between Hilla and Nikka.
Then, you must have that
$$\sum_{r=0}^m f(r) = (m+2)!. \tag1 $$
Assuming that (1) above is satisfied, the expected number of men, between Hilla and Nikka may be computed as
$$\frac{\sum_{r=0}^m \left[r \times f(r)\right]}{(m+2)!}. \tag2 $$
Given Parcly Taxel's answer, which I agree with, then I have reduced the problem to:

*

*Determining a closed form expression for $f(r)$.


*Verifying that the equation in (1) above is satisfied.


*Verifying that the expression in (2) above equals $~\dfrac{m}{3}.$

$\underline{\text{Closed form expression for} ~f(r)}$
Assuming that there are exactly $(r)$ men between Hilla and Nikka, then you have a fused unit of $(r + 2)$ people, and there are exactly $~\displaystyle \binom{m}{r}~$ ways of selecting the $(r)$ men that will be a part of this fused unit.
With Hilla and Nikka required to be on the ends of this fused unit, there are $(2!) \times (r!)$ ways of internally permuting the people inside of this fused unit.
Then, the $(m-r)$ men outside of this fused unit represent other units.  So, you also have $(m-r+1)$ external units to be permuted.
Therefore,
$$f(r) = \binom{m}{r} \times (2!) \times (r!) \times [(m-r + 1)!]$$
$$ = \frac{m!}{r! [(m-r)!]} \times (2!) \times (r!) \times [(m-r + 1)!]$$
$$ = m! \times 2 \times (m-r+1).$$

$\underline{\text{Verification of the equation in (1) 
above }}$
$$\sum_{r=0}^m f(r)$$
$$= \sum_{r=0}^m [ ~m! \times 2 \times (m-r+1) ~]$$
$$= 2(m!) \times \sum_{r=0}^m [ ~(m-r+1) ~]$$
$$= 2(m!) \times \left\{ ~(m+1)^2 - \sum_{r=0}^m [~r~] ~\right\}$$
$$= 2(m!) \times \left\{ ~(m+1)^2 - \frac{m(m+1)}{2} ~\right\}$$
$$= 2(m!) \times (m+1) \times \left\{ ~(m+1) - \frac{m}{2} ~\right\}$$
$$= 2(m!) \times (m+1) \times \left\{ 
~\frac{[ ~2(m+1)~] - m}{2} ~\right\}$$
$$= 2(m!) \times (m+1) \times \left\{ 
~\frac{m + 2}{2} ~\right\}$$
$$= (m+2)!.$$

$\underline{\text{Verification That My Computation Equals} ~\dfrac{m}{3}}$
Temporarily focusing only on the numerator:
$$\sum_{r=0}^m \left[ ~r \times f(r) ~\right]$$
$$= ~\sum_{r=0}^m \left[ ~r \times m! \times 2 \times (m-r+1) ~\right]$$
$$= ~2(m!) \times \left\{ ~\sum_{r=0}^m 
\left[ ~r \times (m-r+1) ~\right] ~\right\}$$
$$= ~2(m!) \times \left\{ ~\sum_{r=0}^m 
\left[ ~\left\langle ~r \times (m+1) ~\right\rangle - r^2 ~\right] ~\right\}$$
$$= ~2(m!) \times \left\{ 
\left[ (m+1) \times \frac{m(m+1)}{2} ~\right] - \frac{m(m+1)(2m+1)}{6} ~\right\}$$
$$= ~2(m!) \times m \times (m+1) \times \left\{ 
\frac{m+1}{2} - \frac{2m+1}{6} ~\right\}$$
$$= ~2(m!) \times m \times (m+1) \times 
\left\{ ~\frac{(3m + 3) - (2m+1)}{6} ~\right\}$$
$$= ~2(m!) \times m \times (m+1) \times 
\left\{ ~\frac{m + 2}{6} ~\right\}$$
$$= ~2[(m+2)!] \times \frac{m}{6}. \tag3 $$
So, the expression in (3) above represents the numerator, and therefore,
$$\frac{\text{the numerator}}{(m+2)!} = \frac{[(m+2)!] \times m}{[(m+2)!] \times 3} = \frac{m}{3},$$
as required.
