# 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

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It wouldn't hurt to observe that $\dfrac{48}{720} = \dfrac{1\cdot48}{15\cdot48}$ and the $48$s cancel, leaving $1/15$, and likewise $\dfrac{12}{180} = \dfrac{1\cdot12}{15\cdot12}$ and the $12$s cancel, again leaving $1/15$. That proves they're exactly equal, whereas $0.06$ is rounded. – Michael Hardy Aug 21 '11 at 23:20

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, $$n(S) = \frac{6!}{2!2!} = 180.$$
As an aside, what are your $E$ and $S$? It seems like $E$ should stand for "events" (i.e. total number of events) and $S$ for "successes", but this is the opposite of how you use them in your question. – Austin Mohr Aug 21 '11 at 18:35