Probability of not making a shoe pair. 
Ten adults enter a room, remove their shoes, and toss their shoes into a pile. Later, a child randomly pairs each left shoe with a right shoe without regard to which shoes belong together. The probability that for every positive integer $k<5$, no collection of $k$ pairs made by the child contains the shoes from exactly $k$ of the adults is $\frac{m}{n}$, where m and n are relatively prime positive integers. Find $m+n.$

I am having an issue understanding the language of the problem. 
I think I need to break into several cases as: $k = 1, k =2, k= 3, k=3$. 
Case 1: ($k = 1$)
no collection of $1$ pair contains shoes from exactly $1$ adult.
Well there are $10$ pairs of shoes there. Ten different designs.
Probability that the pair isnt exactly from $1$ adult is:
$\frac{\binom{20}{1}}{\binom{20}{1}} \cdot \frac{\binom{18}{1}}{\binom{19}{1}} = \frac{18}{19}$
Case 2: $(k=2)$ no collection of $2$ pairs is from $2$ adults.
$$\frac{\binom{20}{1}}{\binom{20}{1}} \cdot \frac{\binom{18}{1}}{\binom{19}{1}}  \cdot \frac{\binom{16}{1}}{\binom{18}{1}} \cdot \frac{\binom{14}{1}}{\binom{17}{1}} =  \frac{80640}{116280} = \frac{2016}{2907}$$
I am not sure what to calculate, the wording is so confusing itself. I think I already have went the wrong path.
Hints are appreciated!
 A: Numbering the right shoes from $1$ to $10$, we can consider that the pairing done by the child is a permutation $\sigma\in S_{10}$. The hypothesis that no $k$ pairs contain the shoes from exactly $k$ adults can be interpreted directly with the decomposition of $\sigma$ in cycles. Actually, we can prove the following result:

For all $k<5$, no set of $k$ pairs contain the shoes of exactly $k$ adults if and only if there are no cycles of size $<5$ in the decomposition of $\sigma$.

Indeed, under the first hypothesis, $\sigma$ cannot have a cycle of size $k<5$ because the corresponding pairs of shoes would form a set of $k$ shoes belonging to exactly $k$ adults.
Conversely, if there is a set $E$ of $k<5$ shoes belonging to exactly $k$ adults, then $\sigma$ contains a cycle composed of elements of $E$, and this cycle is of size $<5$.
Finally, there are only two disjoint cases for $\sigma$:


*

*Either $\sigma$ is a $10$-cycle, which gives us $9!$ possibilities

*Or $\sigma$ is the product of two disjoint $5$-cycles, which gives us $\binom{10}{5}(4!)^2$ possibilities


The probability is therefore
$$\frac{9!+\binom{10}{5}(4!)^2}{10!}=\frac{508032}{3628800}=\frac{7}{50},$$
so that $n+m=57$.
