$k$ socks $n$ drawers Consider the following problem:
There are $k$ socks and $n$ drawers which can only contain 1 sock (very small drawer indeed). The drawers are aligned. You put socks in the drawers with uniform probability. You are wondering the law of probability of the number of drawers containing 1 sock and the drawer to its right have also a sock in it.
More formally, write $[1:n]=[1;n]\cap\mathbb{N}$, and let $\eta$ be a configuration of the sock & drawer, meaning $\eta = (\eta_{i})_{i\in [1:n]}\in \Omega=\{0;1\}^{[1:n]}$ where $\eta_{i}=1$ means that there is a sock in the drawer $i$. Furthermore, denote $\mathbb{P}$ the law of probabilty over $\Omega$ such that  $\forall \eta,\sigma \in \Omega:\mathbb{P}(\eta)=\mathbb{P}(\sigma)$ 
We are looking at the law of probability of the following random variable : $$ \sum_{i=1}^{n-1} \eta_i \eta_{i+1} $$ under the condition $\sum \eta_i=k$
 A: Let $B(n,k)$ be the set of $n$-bit strings with exactly $k$ ones. For $b=b_1b_2\ldots b_n\in B(n,k)$ let 
$$f(b)=|\{i\in[n-1]:b_i=b_{i+1}=1\}|\;;$$
if all we want is the expected value of $f(b)$ when $b$ is chosen from the uniform distribution on $B(n,k)$, we want
$$\binom{n}k^{-1}\sum_{b\in B(n,k)}f(b)\;,$$
which isn’t hard to compute. 
For any $i\in[n-1]$ there are $\binom{n-2}{k-2}$ strings $b=b_1b_2\ldots b_n\in B(n,k)$ such that $b_i=b_{i+1}=1$, so
$$\begin{align*}
\sum_{b\in B(n,k)}f(b)&=\sum_{i\in[n-1]}|\{b\in B(n,k):b_i=b_{i+1}=1\}|\\
&=\sum_{i\in[n-1]}\binom{n-2}{k-2}\\
&=(n-1)\binom{n-2}{k-2}\;,
\end{align*}$$
and hence
$$\binom{n}k^{-1}\sum_{b\in B(n,k)}f(b)=\frac{k!(n-k)!}{n!}\cdot\frac{(n-1)!}{(k-2)!(n-k)!}=\frac{k(k-1)}n\;.$$
If we want the actual distribution of $f(b)$, i.e., 
$$|\{b\in B(n,k):f(b)=m\}|$$
for each $m\in\{0,1,\ldots,k-1\}$, we have to work harder. 
Suppose that $b\in B(n,k)$ has $\ell$ maximal blocks of ones. Each $1$ in a block except the first is the second $1$ of an adjacent pair, so $f(b)=k-\ell$, and $|\{b\in B(n,k):f(b)=m\}|$ is the number of $b\in B(n,k)$ having $k-m$ blocks of ones.
There are $\binom{k-1}{\ell-1}$ ways to divide $k$ ones into $\ell$ non-empty blocks. We now put a $0$ in each of the $\ell-1$ spaces between adjacent blocks of ones; that leaves $n-k-\ell+1$ zeroes to be distributed at will amongst the $\ell+1$ slots before the first block of ones, between adjacent blocks of ones, and after the last block of ones. This can be done in 
$$\binom{(n-k-\ell+1)+(\ell+1)-1}{(n-k-\ell+1)-1}=\binom{n-k+1}{n-k-\ell+1}=\binom{n-k+1}\ell$$
ways, so there are 
$$\binom{k-1}{\ell-1}\binom{n-k+1}\ell$$
strings in $B(n,k)$ with $\ell$ blocks of ones, and
$$\begin{align*}
|\{b\in B(n,k):f(b)=m\}|&=\binom{k-1}{k-m-1}\binom{n-k+1}{k-m}\\
&=\binom{k-1}m\binom{n-k+1}{k-m}\;.
\end{align*}$$
As a quick sanity check note that
$$\sum_m\binom{k-1}m\binom{n-k+1}{k-m}=\binom{n}k\;,$$
as it should.
The probability of choosing a word $b$ such that $f(b)=m$ is therefore
$$\binom{n}k^{-1}\binom{k-1}m\binom{n-k+1}{k-m}\;.$$
