Probability and Permutation with Repetition The problem states:

The letters of the word TOMATO are arranged at random. What is the
  probability that the arrangement begins and ends with T?

I calculate n(S) to be 6! = 720
I calculate n(E) to be 2*4! = 48
I arrive at n(E) thusly:
The number of permutations where "T" is both first and last is 2! = 2. That leaves 4 spaces to fill with the remaining letters O M A O.
To calculate that is 4! = 24  
The probability of the event, then, is $\frac{48}{720}$ = .06
The book's answer says: $\frac{12}{180}$ which also equals .06.  
I'm wondering if I've miscalculated n(S) and n(E) and just serendipitously got the same ratio? Or if my method is correct and the text's answer has skipped directly to the reduced fraction?
Thanks,
n
 A: If you wanted to approach this problem from the view that the T's and O's are indistinguishable, you can fix a T in front and back in exactly 1 way (since we can't tell the difference between the two T's). Now, there are $4!$ arrangements of the remaining 4 letters, but some of these are indistinguishable. To account for this, we divide by the number of ways to permute the O's, which is $2!$. Thus,
$$
n(E) = \frac{4!}{2!} = 12.
$$
For $n(S)$, there would be $6!$ arrangements of the 6 letters, but again some are indistinguishable. We should divide $6!$ by the number of ways to permute the O's and again by the number of ways to permute the T's. Thus,
$$
n(S) = \frac{6!}{2!2!} = 180.
$$
A: The difference probably came about because the book considered the two Os and the two Ts to be indistinguishable, whereas you considered these to be distinct instances of the letters, so each of their arrangements corresponds to four of your arrangements. Your approach is slightly better, since you didn't have to worry about dividing out the number of equivalent arrangements, which is unnecessary in this case.
