Question

a classic counting problem is to determine the number of different ways that the letters of \possession\ 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 \possession\ can be arranged is (type an integer or a simplified fraction.)

Explanation:

Step1: Count letter - counts

The word "possession" has 10 letters. There are 4 's's, 2 'o's, 2 's's (already counted in the 4 's's for simplicity), 1 'p', 1 'e', 1 'i'.

Step2: Use permutation formula for multi - sets

The formula for permutations of a multi - set with \(n\) objects, where there are \(n_1\) of one kind, \(n_2\) of another kind, \(\cdots\), \(n_k\) of the \(k\)th kind is \(\frac{n!}{n_1!n_2!\cdots n_k!}\). Here \(n = 10\), \(n_1=4\) (for 's') and \(n_2 = 2\) (for 'o'). So the number of arrangements is \(\frac{10!}{4!2!}\).
\[

$$\begin{align*} 10!&=10\times9\times8\times7\times6\times5\times4\times3\times2\times1 = 3628800\\ 4!&=4\times3\times2\times1= 24\\ 2!&=2\times1 = 2\\ \frac{10!}{4!2!}&=\frac{3628800}{24\times2}\\ &=\frac{3628800}{48}\\ & = 75600 \end{align*}$$

\]

Answer:

75600