The first part is right.

I don't think the second part is right though. The probability of choosing an M as the first letter is going to be 1choose(12) = n!/(n-1)!k!

So,

1choose(12) = 1!/11!12!

Now we want to choose an S as the last letter. The probability is now 1choose11 since we have already chosen an M and there are 1 fewer letters to choose. So,

1choose(11) = 1!/10!11!

Since we want both of these events to happen we multiply them together. To get the probability of this happening we divide by the total number of possibilities which you have calculated in the first part. So,

(1choose(12) * 1choose(11)) / (12!/(2!2!2!))

Hopefully this makes sense.