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