How to calculate the hybridization of ethane with H=½[V+M-C+A]? I wanted to calculate the hybridization of this molecule $\ce{CH3-CH3}$. I noticed a shortcut formula given in my class notes for calculating the hybridization. It was 

$$H=\frac{1}{2}×[V+M-C+A].$$

With 


$H$: Hybridization 
$V$: Number of valence electrons of central atom.
$M$: Number of monovalent atoms (i.e hydrogen or halogen) around central atom.
$C$: Cationic Charge.
$A$: Anionic Charge.


When I tried to calculate the hybridisation of $\ce{CH3-CH3}$ by using above formula, where for any one of the carbon atom $V=4$, $M=3$, $C=0$, $A=0$ then $H=\frac{7}{2}$. But as we all know hybridization of every carbon in ethane is $\ce{sp^3}$, i.e. $H$ should be $4$. I know other methods of calculating $H$ like $H= \text{no. of sigma bonds} + \text{no.of lone pairs}$, which gives me the right answer, but why does the above formula give something different? And also for any molecule if using that formula gives fractional answer then what does that mean?
The mistake that you have made is incorrect understanding of M. From what I remember, M was number of single bonds, which would be only with a monovalent atom, in the context you learnt tbe formula. This was taught in chemical bonding where only atoms are bonded to central atom. 
Ex- PCl5, PCl3  
You should modify this definition to get the answer  here. Any monovalent group (-CH3) should be there in M.