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 7×10!/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 2×10!/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 3×10!/2!2!2!
Adding these two cases gives me 7×10!/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?
Best Answer
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.