Hmm! Chris, suppose there are 4 volunteers where each subgroup consists of 1 participant.
If there is one subgroup then clearly there are 4 possible choices.
However, if there are 4 subgroups, it seems to me there is only one way to do this; but your method would give nCr(4,1)*nCr(3,1)*nCr(2,1) = 24 ways!! This would only be correct if each subgroup were given a different label and it was the number of combinations of label and participant that was required. Perhaps the subgroups are given labels - this would be quite sensible. However, it isn't clear to me that that was what was intended here.