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: Identify repeated letters

The word "possession" has 10 letters. There are 2 'p's, 4 's's, 2 'o's.

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 \(n = 10\), \(n_1=2\) (for 'p'), \(n_2 = 4\) (for's'), \(n_3=2\) (for 'o'). So the number of arrangements is \(\frac{10!}{2!4!2!}\).

Step3: Calculate factorial values

We know that \(n!=n\times(n - 1)\times\cdots\times1\). So \(10!=10\times9\times8\times7\times6\times5\times4\times3\times2\times1 = 3628800\), \(2!=2\times1=2\), \(4!=4\times3\times2\times1 = 24\). Then \(\frac{10!}{2!4!2!}=\frac{3628800}{2\times24\times2}=37800\).

Answer:

37800