Ten children are playing a "untangling" game: they stand in a circle, and with their eyes closed and arms outstretched, they start moving towards the center of the circle. Everyone grabs someone else's hand with both of their hands. Then they open their eyes and start "untangling": they duck under each other's hands, step over each other's hands, etc. (everyone is sufficiently flexible) - but they do not let go of each other's hands in the meantime. In what percentage of all cases is it true that if two adjacent children let go of each other's hands at one point, the ten children holding each other's hands form a connected chain? (H)
The last sentence's condition means that the 10 children formed a single "circle" before two of them let go of each other's hands. The complex formulation is due to the possibility of a knot or self-intersection on this circle. We will refer to any arrangement that meets the condition, even if it contains a knot, as a "circle". We claim that $n$ children can form a "circle" in $(2 n-2)(2 n-4) \cdots \cdot 4 \cdot 2$ ways.

To prove our claim, consider one of the children, say Pista, as the starting point. He can hold hands with $(2 n-2)$ other hands (since the condition implies that no one can hold their own hand). The child whose hand he holds can then hold hands with $(2 n-4)$ other hands (holding Pista's other hand would close the circle). Continuing this way, the second-to-last child can choose from the two hands of the last child, who will then hold Pista's other hand. This results in $(2 n-2)(2 n-4) \cdots \cdot 2$ possibilities, which we denote as $(2 n-2)!!$ and call the semi-factorial of $2 n-2$.

For $n=10$, this is 18!!. If the ten children do not form a single "circle", they will form smaller "circles" (the relative positions of which do not concern us). The sizes of these circles can vary from 2 to 8. The number of arrangements for given sizes is obtained by dividing the children into groups of the given sizes, then independently forming the possible circles within each group.

For example, if two 3-member and one 4-member circle are formed, we first

$$
\binom{10}{3} \cdot\binom{7}{3} \cdot\binom{4}{4}=\binom{10}{3}\binom{7}{3} \text { ways }
$$

arrange the children, then within each group, we form the circles in $(6-2)!!,(6-2)!!$, and $(8-2)!!$ ways, respectively. However, in this arrangement, we distinguish the two 3-member groups - and generally the groups of the same size, so here we count each possibility twice. Similarly, the (2, 2, 2, 4) group arrangements, for example, are counted $3!=6$ times, and the (2,2,3,3) arrangements are counted $2!\cdot 2!$ times. Taking all these into account, the following possibilities arise. At the beginning of the list, we have listed the sizes of the groups in the given partition, which is all possible representations of 10 as the sum of numbers between 2 and 8.

$$
\begin{aligned}
& 2)!!=1857945602,8 ;\binom{10}{2} \cdot(4-2)!!\cdot(16-2)!!=580608003,7 ;\binom{10}{3} \cdot(6-2)!!\cdot(14-2)!!=442368005 \\
& 5 ;\binom{10}{5} \cdot(10-2)!!\cdot(10-2)!!/ 2=18579456 \\
& \text { because we counted each case twice. } 2,2, \\
& 6 ;\binom{10}{6}\binom{4}{2} \cdot(4-2)!!\cdot(4-2)!!\cdot(12-2)!!/ 2=96768002,3,5 ;\binom{10}{5}\binom{5}{3} \cdot(4-2)!!\cdot(6-2)!!\cdot(10-2)!!=154828802 \\
& 4,4 ;\binom{10}{2}\binom{8}{4} \cdot(4-2)!!\cdot(8-2)!!\cdot(8-2)!!/ 2=72576003,3,4 ;\binom{10}{4}\binom{6}{3} \cdot(6-2)!!\cdot(6-2)!!\cdot(8-2)!!/ 2=64512002 \\
& 2,2,4 ;\binom{10}{4}\binom{6}{2}\binom{4}{2} \cdot(4-2)!!\cdot(4-2)!!\cdot(4-2)!!\cdot(8-2)!!/ 3!=12096002,2,3,3 ;\binom{10}{3}\binom{7}{3}\binom{4}{2} \cdot(4-2)!!\cdot(4- \\
& 2)!!\cdot(6-2)!!\cdot(6-2)!!/(2 \cdot 2)=16128002,2,2,2,2 ;\binom{10}{2}\binom{8}{2}\binom{6}{2}\binom{4}{2} \cdot((4-2)!!)^{5} / 5!=30240
\end{aligned}
$$

The sum of these gives the total number of possible cases, which is 387099 936. The desired ratio is $\frac{185794560}{387099936} \approx 0.48$, so in slightly less than half of the cases, the ten children form a single "circle".

Remark. If $N$ children are to be arranged into $m_{1}$ circles of size $c_{1}, m_{2}$ circles of size $c_{2}, \ldots, m_{k}$ circles of size $c_{k}$ ($\left(m_{1} c_{1}+\right.$ $m_{2} c_{2}+\cdots+m_{k} c_{k}=N$ ), then this is possible in

$$
\frac{N!}{c_{1}^{m_{1}} \cdot c_{2}!!^{m_{2}} \cdots c_{k}!^{m_{k}}} \cdot \frac{1}{m_{1}!\cdot m_{2}!\cdots m_{k}!} \cdot\left(c_{1}!!\right)^{m_{1}} \cdot\left(c_{2}!!\right)^{m_{2}} \cdots\left(c_{k}!!\right)^{m_{k}}
$$

ways.