Counting Couples A group of m men and w women randomly sit in a single row at a theater. If a man and woman are seated next to each other, they form a "couple." "Couples" can overlap, which means one person can be a member of two "couples." 
Question: What is the expected number of couples?
Comment:
I have a hard time with word problems that deal with "expectations". 
 A: Consider a woman. If she’s sitting at one end of the row, there are $m(m+w-2)!$ permutations in which she is part of a couple. Otherwise, there are $m(m-1)(m+w-3)!$ permutations in which she is part of two couples and $2m(w-1)(m+w-3)!$ permutations in which she is part of one couple. There are $(m+w-1)!$ permutations with her in that seat, so the expected number of couples containing a randomly chosen woman is
$$\frac2{m+w}\cdot\frac{m(m+w-2)!}{(m+w-1)!}+\frac{m+w-2}{m+w}\cdot\frac{2m(m+w-3)!(m+w-2)}{(m+w-1)!}\;.$$
This simplifies to
$$\frac{2m(m+w-1)}{(m+w)(m+w-1)}=\frac{2m}{m+w}\;.$$
There are $w$ women, so by linearity of expectation the expected number of couples is
$$\frac{2mw}{m+w}\;.$$
(As a partial check against obvious errors, notice that this expression is unchanged if we interchange the rôles of $m$ and $w$, as it obviously should be.)
A: Let $n=m+w$. Let $X$ be the number of couples, a random variable. Let Irma be one of the women and let Jack be one of the men. There are $n(n-1)$ ways that Irma and Jack can be seated, and there are $2(n-1)$ ways they can be seated next to each other. The probability that Irma and Jack are seated next to each other is $\frac{2(n-1)}{n(n-1}=\frac 2 n$, and this is the expected value of the indicator variable $X_{\text{Irma,Jack}}$ whose value is $1$ if Irma and Jack are seated next to each other, $0$ otherwise. Summing over all $mw$ potential couples, $E[X]=\frac{2mw}n=\frac{2mw}{m+w}$.
A: This problem may also be solved using generating functions. Let $p_{n,k}$ be the number of sequences of length $n$ with $k$ couples where the last person is male. Similarly, let $q_{n,k}$ be the number of sequences of length $n$ with $k$ couples where the last person is female.
This gives the following relations:
$$ p_{n,0} = u^n \\ q_{n,0} = v^n \\
p_{n,k} = u z \times q_{n-1,k-1} + u \times p_{n-1,k} \\
q_{n,k} = v z \times p_{n-1,k-1} + v \times q_{n-1,k}.$$
Here the variable $z$ represents the number of couples.
Now introduce $$P(z, u, v) = \sum_{n\ge 0} \sum_{k\ge 0} p_{n, k} \\
Q(z, u, v) = \sum_{n\ge 0} \sum_{k\ge 0} q_{n, k}.$$ 
Summing the recurrence relations, we find
$$  \sum_{n\ge 1} \sum_{k\ge 1} p_{n,k} = P
- \sum_{k\ge 1} p_{0, k} - \sum_{n\ge 1} p_{n, 0} =
P - \frac{u}{1-u} \\ =
u z \times Q + u \times \left(P - \sum_{n\ge 0} p_{n, 0}\right)
= u z \times Q + u \times \left(P - \frac{u}{1-u}\right)$$
and similarly
$$  \sum_{n\ge 1} \sum_{k\ge 1} q_{n,k} = Q
- \sum_{k\ge 1} q_{0, k} - \sum_{n\ge 1} q_{n, 0} =
Q - \frac{v}{1-v} \\ =
v z \times P + v \times \left(Q - \sum_{n\ge 0} q_{n, 0}\right)
= v z \times P + v \times \left(Q - \frac{v}{1-v}\right).$$
Solving this system of equations, we obtain
$$ P =  -{\frac { \left( vz-v+1 \right) u}{v-1-uv+u+u{z}^{2}v}}
\quad \text{and} \quad
Q = -{\frac {v \left( -u+1+uz \right) }{v-1-uv+u+u{z}^{2}v}},$$
so that
$$ P+Q = -{\frac {2\,uzv-2\,uv+u+v}{v-1-uv+u+u{z}^{2}v}}.$$
Now observe that
$$ \left.(P+Q)\right|_{z=1} = -{\frac {u+v}{v-1+u}} =
\sum_{q\ge 1} (u+v)^q,$$
and
$$ [u^m v^w] \sum_{q\ge 1} (u+v)^q = \binom{m+w}{m}.$$
This is a nice sanity check, a kind of hash certificate that shows that we have the right generating function.
To conclude note that
$$ \left.\left(\frac{d}{dz} (P+Q)\right)\right|_{z=1} = 
2\,{\frac {uv}{ \left( v-1+u \right) ^{2}}}$$
so that
$$ [u^m v^w] 2 uv \sum_{q\ge 0} (q+1) (u+v)^q
= 2 [u^{m-1} v^{w-1}] \sum_{q\ge 0} (q+1) (u+v)^q =
2 (m+w-1)\binom{m+w-2}{m-1}.$$
It follows that the expected number of couples $E[C]$ is
$$ E[C] = \frac{2 (m+w-1)\binom{m+w-2}{m-1}}{\binom{m+w}{m}} =
\frac{2mw}{m+w}.$$
I do believe that this is a nice exercise in the use of ordinary generating functions including the sanity check (obviously the number of arrangements is $m+w$ choose $m$.) With these generating functions we can calculate arbitrary factorial moments of $C.$ 
A: Continuing the computation we can calculate $E[C(C-1)]$.
We have $$\left.\left(\left(\frac{d}{dz}\right)^2 (P+Q)\right)\right|_{z=1} =
2\,{\frac {uv \left( -2\,uv-u+{u}^{2}+{v}^{2}-v \right) }{ \left( v-1+u \right) ^{3}}}.$$
After a straightforward calculation this transforms into
$$ E[C(C-1)] = 
{\frac {2\,mw \,\left( 2\,mw-w-m \right) }{ \left( m+w-1 \right)  \left( m+w \right) }}.$$
