In HCP unit cell number of effective number of atoms is $6$,

so  $A=6$.

Number of tetrahedral voids

                                  $=$ 2 x number of atoms forming the lattice 

                                  $=$$2 \mathrm{X} 6$

${\mathrm{M}}_{12\times \frac{2}{3}}{ \mathrm{A}}_6$
${\mathrm{M}}_8{ \mathrm{A}}_6$
${\mathrm{M}}_4{ \mathrm{A}}_3$