Probability: Shoes in a row 
20 shoes, from 10 pairs of shoes, are lined randomly. What is the
probability that there is a set of 10 consecutive shoes with 5 shoes
left and 5 right shoes?

I thought that would be a good idea imagine 10 shoes as one unique object, as follow:
Instead enumerate 20 objects, let's separate 10 shoes and think them as another object (call it X), so that there is 11 in total.
There is 11! ways to permute each objects. The problem is to know how to count X. I mean, the first way i thought was: $|X| = \frac{10!}{5!}\frac{10!}{5!}$, so that the answer would be: $\frac{\frac{10!}{5!}\frac{10!}{5!}*11!}{20!}$. But so i realized that |X| that i counted was wrong, what actually i counted for |X| was counting first the 5 possible right shoes, and after this, the 5 possible let shoes.
So i am stuck in how to count |X| right. I thought that we could first choose $C_{10,5} = 10!/(5!5!)$ right shoes, $C_{10,5}$ left shoes. Now the probability for one aleatory choice of X is: $\frac{\frac{10!}{5!}*11!}{20!}$, but X have more than one option, so the probability will be the sum of all possibility for X,
namely $C_{10,5}*C_{10,5}*\frac{{10!}*11!}{20!}$. But this is greater than 1 (...), i have no idea what to do now.
 A: The probability is $1$.
Label the shoes from $1$ to $20$ according to their position. For $n \in \lbrace 1, \dotsc, 20 \rbrace$, set $$\varepsilon_{n} = \begin{cases} 0 & \text{if the } n \text{th shoe is left}\\ 1 & \text{if the } n \text{th shoe is right} \end{cases} \, \text{.}$$ For $N \in \lbrace 1, \dotsc, 11 \rbrace$, define $$\ell_{N} = \sum_{n = N}^{N +9} \varepsilon_{n} \in \lbrace 0, \dotsc, 10 \rbrace$$ to be the number of left shoes whose label is in $\lbrace N, \dotsc, N +9 \rbrace$.
I pretend that there exists $N_{0} \in \lbrace 1, \dotsc, 11 \rbrace$ such that $\ell_{N_{0}} = 5$. We have $$\ell_{1} +\ell_{11} = \sum_{n = 1}^{20} \varepsilon_{n} = 10 \, \text{,}$$ and hence we have $$\ell_{1} \leq 5 \leq \ell_{11} \quad \text{or} \quad \ell_{11} \leq 5 \leq \ell_{1} \, \text{.}$$ Now, note that, for all $N \in \lbrace 1, \dotsc, 10 \rbrace$, we have $$\ell_{N +1} = \ell_{N} +\varepsilon_{N +10} -\varepsilon_{N} \in \left\lbrace \ell_{N} -1, \ell_{N}, \ell_{N} +1 \right\rbrace \, \text{.}$$ Therefore, by some kind of discrete intermediate value theorem, there exists $N_{0} \in \lbrace 1, \dotsc, 11 \rbrace$ such that $\ell_{N_{0}} = 5$. More precisely, we can take $$N_{0} = \begin{cases} \max\left\lbrace N \in \lbrace 1, \dotsc, 11 \rbrace : \ell_{N} \leq 5 \right\rbrace & \text{if } \ell_{1} \leq 5 \leq \ell_{11}\\ \min\left\lbrace N \in \lbrace 1, \dotsc, 11 \rbrace : \ell_{N} \leq 5 \right\rbrace & \text{if } \ell_{11} \leq 5 \leq \ell_{1} \end{cases} \, \text{.}$$
