What is the sum of radial and angular nodes in the following orbitals of H-atom?

$\left(\mathrm{I}\right) {\mathrm{\psi }}_{2{\mathrm{p}}_{\mathrm{x}}} \left(\mathrm{II}\right) {\mathrm{\psi }}_{2s} \left(\mathrm{III}\right) {\mathrm{\psi }}_{3{\mathrm{d}}_{\mathrm{x}}} \left(\mathrm{IV}\right) {{\mathrm{\psi }}_{3\mathrm{d}}}_{{\mathrm{x}}^2-{\mathrm{y}}^2}$

Radial nodes = n – $\mathcal{l}$ – 1

$\Rightarrow$ (I) n = 2,  $\mathcal{l}$= 1, radial node = 2 – 1 – 1 = 0
Angular node = 1 (YZ plane)
(II) n = 2, $\mathcal{l}$ = 0, Radial node = 2 – 0 – 1 = 1
Angular node = 0
(III) n = 3, $\mathcal{l}$ = 2, Radial node = 3 – 2 – 1 = 0, and

Angular node = 2 (XY and YZ planes)
(IV) n = 3,  $\mathcal{l}$= 2, Radial node = 0, and
Angular nodes = 2
Total = I + II + III + IV = 1 + 1 + 2 + 2 = 6