If the first consonant is $M$ or $T$ and $A,(T/M)$ are repeated :
$2\times\frac{10!}{2!2!}=\frac{10!}{2!}$
If the first consonant is $C,H$ or $S$ and $A,T,M$ are repeated:
$3\times\frac{10!}{2!2!2!}$
Thus total amount equals:
$$\frac{10!}{2!}+3\times\frac{10!}{2!2!2!}=\frac{2!2!10!+3\times10!}{2!2!2!}=\frac{4\times10!+3\times10!}{8}=7\times\frac{10!}{8}$$
We can ensure this result with a reversal case. Choosing a vocal as the first letter, total amount of choices is:
$$\frac{10!}{2!2!}+2\times\frac{10!}{2!2!2!}=\frac{10!}{2!}=\frac{4\times10!}{8}$$
Thus total amount of combinations starting with either consonant or vocal is
$$7\times\frac{10!}{8}+4\times\frac{10!}{8}=\frac{11!}{2!2!2!}=\binom{11}{2,2,2}$$
which equals every permutation of the word, when duplicate letters are concerned.
using 9 letters the number of different words is $\binom{9!}{4!2!}$
to find how many words you can form with 8 letters you have to perform the same calculation on each of the following set of letters
EEEVRRGN, EEEERRGN, EEEEVRGN, EEEEVRRN, EEEEVRRG
e.g. for the first one the number of words is $\binom{8!}{3!2!}$, and so on
then you have to add to the previous results the same calculation for 7 letters considering all the different sets, and precisely
EEVRRGN, EEERRGN, EEEVRGN, EEEVRRN, EEEVRRG, EEEERGN,EEEERRN,EEEERRG, EEEEVGN, EEEEVRN,EEEEVRG, EEEEVRR
I don't know whether there exists a different direct way
Best Answer
The purpose of dividing is to reverse a multiplication which has taken place, which should not have taken place.
Generally, the multiplication "should not" have taken place because two outcomes are not "distinct", i.e. they are the same.
For example if I ask how many distinct ways two coins can turn out, you will say there are 2 ways for the first coin, and 2 ways for the second so the answer is $2\times2=4$.
However if I then tell you that one of the coins is a double-headed coin, you now know that the 2nd multiplication by 2 is in error, as there is only one distinct way that coin can turn out. So you divide by $2$ to give your answer $\frac{4}{2}=2$.
In this example it's obvious the answer is two but sometimes when you write out a solution it's easier to multiply all the possibilities first and then divide to "uncount" the ones that aren't distinct.