A classic counting problem is to determine the number of different ways that the letters of "occasionally" can be arranged. Find that number. If th that the letters will be in alphabetical order? The number of different ways that the letters of "occasionally" can be arranged is \( 29,937,600 \) (Type an integer or a simplified fraction.) The probability that the letters will appear in alphabetical order is (Type an integer or a simplified fraction.)

To determine the number of different arrangements of the letters in "occasionally," we need to account for the fact that the word contains repeated letters. The letters are: o, c, c, a, s, i, o, n, a, l, l, y. Counting the repetitions, we have: - 2 'o's - 2 'c's - 2 'a's - 2 'l's Using the formula for permutations of a multiset: \[ \text{Total arrangements} = \frac{n!}{n_1! \cdot n_2! \cdot ... \cdot n_k!} \] \(n = 12\) (total letters), and \(n_1, n_2, ...\) are the frequencies of each letter. \[ = \frac{12!}{2! \times 2! \times 2! \times 2!} = \frac{479,001,600}{16} = 29,937,600. \] The probability of the letters being in alphabetical order is determined by considering that there is only one specific arrangement of the letters that represents alphabetical order (a, a, c, c, i, l, l, n, o, o, s, y) versus the total arrangements (29,937,600): \[ \text{Probability} = \frac{1}{29,937,600}. \]