Is there a complete set of mutually disjoint trios in a (5,8,24) Steiner system? Question: In a (5,8,24) Steiner system, is it possible to find a set of 759/3=253 trios such that every octad lies in exactly one of these trios?
Background information: 
If $X$ is a set of size 24, then a (5,8,24) Steiner system is a set of subsets of $X$ of size 8, called octads, such that every subset of $X$ of size 5 is contained in a unique octad.  It is known that a (5,8,24) Steiner system is unique up to a permutation of $X$.  Fix such a Steiner system $S$.
It is easy to calculate that there are $\binom{24}{5}/\binom{8}{5} = 759$ octads in $S$.  A trio is a set of 3 mutually disjoint octads covering $X$.  It is known that every octad lies in exactly 15 trios and hence there are $759 \cdot 15 / 3 = 3795$ trios in $S$.
 A: Yes, such a family can be constructed. Let us begin by constructing the Steiner system as supports of words of weight 8 in the extended Golay code. Here the $[23,12,7]$ binary Golay code $G$ is the ideal $G:=R\cdot p$ of $R=F_2[x]/\langle x^{23}+1\rangle$ generated by the factor $$p(x)=1+x+x^5+x^6+x^7+x^9+x^{11}\mid x^{23}+1.$$ We view polynomials as strings of 23 bits simply by listing their coefficients, and index bit positions with the exponents. The code $G$ is extended to a code $\overline{G}$ of length 24 by adding an overall parity bit to the end - call this last position $\infty$ with a view of matching the bit positions with points of the projective line over $F_{23}$. So the polynomial $p(x)$ corresponds to a word of weight 7 in $G$, and in the extended code $\overline{G}$ it becomes a word of weight 8 giving the octad $o_1=\{0,1,5,6,7,9,11,\infty\}$.
Being an ideal $G$ is stable under multiplication by $x$, i.e. the 23-cycle 
$\alpha(i)\equiv i+1\pmod{23}$ (acting on the bit positions or the set of exponents of $x$, whichever way you prefer). Being an ideal $G$ is also stable under the Frobenius map = squaring. As the order of $2$ modulo $23$ is equal to $11$ this breaks the bit positions/exponents into two 11-cycles:
$$\beta=(0)(1,2,4,8,16,9,18,13,3,6,12)(5,10,20,17,11,22,21,19,15,7,14)(\infty)$$ 
or $\beta(i)\equiv2i\pmod{23}$. We view both of these as permutations in $Aut(\overline{G})\cong M_{24}$. They are both in the point stabilizer of $\infty$.
We see that $\beta\alpha\beta^{-1}=\alpha^2$ ($i\mapsto i+2\pmod{23}$). Therefore $\beta$ normalizes the group $\langle\alpha\rangle$, and together they generate a group $H=
\langle \alpha,\beta\rangle\cong C_{23}\rtimes C_{11}$ of size $11\cdot23=253$.
It is easy to see that $H$ does not fix any octads. This is because $M_{24}$ acts transitively on the set of $759=3\cdot11\cdot23$ octads, and neither $11^2$ nor $23^2$ divide the order $|M_{24}|$. Therefore the octads are partitioned into three full size orbits of $H$.
With the aid of CAS (I used Mathematica) it is easy to generate all the octads and list all those that have supports disjoint from $o_1$. There are exactly 30 of those (15 pairs, as predicted by your data telling that $o_1$ is a member of exactly 15 trios). The method that I used in generating them spewed out $o_2=\{3,4,8,10,16,19,21,22\}$ as the first octad of this kind. Because $H$ stabilizes $\infty$ it is immediately clear that $o_2$ does
not belong to the orbit $H\cdot o_1$. It is straightforward to check that the octad complementing this trio $o_3=\{2,12,13,14,15,17,18,20\}$ does not belong to the orbit $H\cdot o_2$. Thus all the octads belong to exactly one of the orbits $H\cdot o_j, j=1,2,3$. For any $\gamma\in H\le M_{24}$, the octads $\gamma(o_j),j=1,2,3$ obviously form a trio. Thus together these 253 trios cover all the octads as requested.
This solution is admittedly somewhat unsatisfactory in the sense that at a critical point, $o_3\notin H\cdot o_2$, I used brute force.  It would not surprise me, if the same thing happened to any trio containing $o_1$. I verified this for one other pair of octads $o_2',o_3'$ completing $o_1$ to a trio. The simple idea of using the group $H$ is my key input. After finding $o_2$ the rest could probably be done without the aid of a computer. After all, the group $H$ consists of affine mappings from $F_{23}\cup\{\infty\}$ to itself of the form $i\mapsto ui+v$, where $u$ is a quadratic residue modulo $23$ and $v\in F_{23}$ is arbitrary.
Edit: Ted already noticed that I had jumped to a 'conjecture' on too scant data. Today I checked out all the 15 trios including $o_1$. The exhaustive tally is that 9 out of those work the same way as the example trio $\{o_1,o_2,o_3\}$. For the remaining 6 trios the two complementary octads actually belong to the same orbit of $H$, and hence those trios cannot be used to solve Ted's question. At least not using this particular conjugate of $H$ (another one may work).
