How to find the number of permutations of the letters of the word MATHEMATICS that begin with a consonant I have to find the number of permutations of letters of the word "MATHEMATICS" which start with consonants only.
My attempts:
Attempt 1:
There are 7 consonants (including the repeated M&T), therefore, there are 7 ways of filling in the first letter. Then the rest can be filled in 10! ways. Since A,T and M are repeated, I'll divide by 2!2!2! giving me 7x10!/6
Attempt 2:
Case 1: M or T are starting letters. Let's say I pick M. Then I can arrange the other letters in 10! ways. I'll divide by 2! for the two T's, and by another 2! for the two A's. I could do the same thing if I picked T for the starting letter. Therefore I get 2x10!/2!2!
Case 2: H,C,S are the starting letters. Therefore I have 3 ways of filling in the first place. 10! for the rest and I divide by 2!2!2! For the repetition of M,T,A giving 3x10!/2!2!2!
Adding these two cases gives me 7x10!/8
(The answer given is 11!/8 which I think is incorrect since that is the number of ways even if vowels would be allowed to be the first letter. )
Is any of my attempts correct? If so why is the other one wrong? If both are wrong, why?
 A: 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.
A: I would like to direct you to the answer I just gave here
As a brief explanation the general technique applied to your problem: 


*

*Find the number of ways to choose a consonant to put in one place (call this $k_1$).

*Take this out of your collection and put it where it needs to be.

*Find the number of ways you can permute the remaining $n-1$ objects in your collection without the restriction since that is already taken care of (call this $k_2$).

*The final solution is $k=k_1k_2$

