expected value of getting couples on adjoining seats for b bachelors and m models in the n seats row where n=m+b I am analyzing the The Theater Row Puzzle from the book Fifty Challenging Problems in Probability by Mosteller.

"Eight eligible bachelors and seven beautiful models
  happen randomly to have purchased single seats in the same 15-seat row
  of a theater. On the average, how many pairs of adjacent seats are
  ticketed for marriageable couples?"

I understand the first part of the solution where the probability of the getting a couple in the first two seats equals to $$\frac{8}{15} \frac{7}{14} + \frac{7}{15} \frac{8}{14} = \frac{8}{15}$$

the prresented solution for b bachelors and m models: 

$$(m+b-1) \ \big{[}\frac{bm}{(m+b)(m+b-1)} + \frac{mb}{(m+b)(m+b-1)}\big{]}=\frac{2mb}{m+b}$$
for b=8 and m=7
$$\frac{2*7*8}{8+7}=7\frac{7}{15}=7.4666...$$

However I don't get the logic for the next seats. Why the probability for getting coupoles on the two first seats is just multiplied by number of seats minus one? 
 Shouldn't the number of bachelors and models be decreasing with every next seat? 

Doesn't the multiplication by (m+b-1) assume that number of bachelors and models are the same for every seat. 

I don't get this. Can anybody clarify the solution?  
 A: Expectation is Linear.   Contrary to the intuition of many, many, new students, independence or otherwise of the random variables being summed does not affect this.
We do, however, make use of the fact that the first $m+b-1$ consecutive pairs  of seats will have identical (marginal) distribution for eligibility.
You are calculating the expectation of the sum of indicator random variables that a seat and the subsequent seat holds complementary sexed individuals, summed for each of the first $m+b-1$ seats (because the last seat does not have a subsequent seat.).
$$\begin{align}\mathsf E(C) &= \mathsf E(\sum_{j=1}^{m+b-1}\mathbf 1_{\text{seats $j$ and $j+1$ are eligible}}) \\ &= \sum_{j=1}^{m+b-1}\mathsf E(\mathbf 1_{\text{seats $j$ and $j+1$ are eligible}}) \\ &= \sum_{j=1}^{m+b-1}\mathsf P({\text{seats $j$ and $j+1$ are eligible}}) & \text{by definition of indicator r.v.} \\ &= (m+b-1)~\mathsf P(\text{seats $1$ and $2$ are eligible})  & \text{since identically distributed}\\ &= (m+b-1)\frac{2mb}{(m+b)(m+b-1)}\\ &= \dfrac{2mb}{m+b}\end{align}$$
