NMR of trigonal bipyramidal molecules I'm slightly confused about NMR.  

I seem to remember at school that if a molecule is symmetrical in $\ce{^1H}$ NMR, that only one peak would be shown e.g. propane's terminal $\ce{-CH3}$. 

Consider the two molecules shown below. 

My professor remarked that both these isomers would show a doublet and triplet in their proton-decoupled $\ce{^31P}$ NMR and I was struggling to work out why. 

Consider the molecule on the left as molecule A and the right as molecule B.

My train of logic was that the axial $\ce{Ph3P}$ group on molecule A would be split by each $\ce{P}$ atom that is equatorial, so using the expression $2nI + 1$ where $I = \frac{1}{2}$ so our singlet would be split into a triplet. 

Likewise, the equatorial $\ce{Ph3P}$ groups would both only see one axial $\ce{Ph3P}$ group, hence only a singlet would be observed. 

I used the same logic on molecule B. 

Is this the correct reasoning for the doublet and triplets observed in both molecules?
“Likewise the equatorial Ph3P groups would both only see one axial Ph3P group, hence only a singlet would be observed” - I think you’ve just made a little mistake. The equatorial groups can see one phosphorus other than themselves, so (2nI + 1) = 2, not 1!