= Solution
Use <Wade's rules> with the metal included as a skeletal vertex. Each BH vertex supplies two skeletal <electrons>, each CH supplies three, and the two additional <hydrogen> atoms supply two more. The $\mathrm{Ru(CO)_3}$ fragment supplies $8+3(2)-12=2$ skeletal <electrons> after the metal's non-skeletal orbitals have been accounted for. Thus
$$
N_{\mathrm{skeletal}}=3(2)+2(3)+2+2=16,\qquad \mathrm{SEP}=8=6+2.
$$
The six-vertex cluster is therefore \b[nido, with a pentagonal-pyramidal skeleton]: Ru is the apex, bonded to the five vertices of the planar $\mathrm{B_3C_2}$ base, and the three <carbonyl ligands> project away from that base. This is the six-vertex fragment obtained by removing an equatorial vertex from a seven-vertex <pentagonal bipyramid>. The five ring atoms carry their terminal B–H or C–H bonds; the two additional <hydrogens> belong to the hydrogen-rich open-face bonding framework. The skeletal <electron> count does not, by itself, locate every <bridging hydride>, so a unique detailed <hydrogen> arrangement should not be inferred from it.
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