Question

a classic counting problem is to determine the number of different ways that the letters of \generally\ can be arranged. find that number. if the letters are mixed up in a random sequence, what is the probability that the letters will be in alphabetical order?
the number of different ways that the letters of \generally\ can be arranged is 90,720. (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.)

Explanation:

Step1: Calculate total arrangements

The word "generally" has 9 letters. But the letter 'e' appears 3 times and the letter 'l' appears 2 times. The formula for permutations of a word with repeated letters is $\frac{n!}{n_1!n_2!\cdots n_k!}$, where $n$ is the total number of letters and $n_i$ are the number of times each repeated - letter appears. Here, $n = 9$, $n_1=3$ (for 'e') and $n_2 = 2$ (for 'l'). So the number of arrangements is $\frac{9!}{3!2!}=\frac{9\times8\times7\times6\times5\times4\times3!}{3!\times2\times1}=90720$.

Step2: Calculate probability of alphabetical order

There is only 1 way for the letters to be in alphabetical order out of the total number of arrangements. So the probability $P=\frac{1}{90720}$.

Answer:

The probability that the letters will appear in alphabetical order is $\frac{1}{90720}$