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 (type an integer or a simplified fraction.)

Explanation:

Step1: Count total letters and repeated letters

The word "generally" has 9 letters. The letter 'e' appears 3 times and the letter 'l' appears 2 times.

Step2: Use permutation formula for multi - sets

The formula for permutations of a multi - set with \(n\) objects where \(n_1\) 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 \(n = 9\), \(n_1=3\) (for 'e') and \(n_2 = 2\) (for 'l'). So the number of arrangements is \(\frac{9!}{3!2!}\).

Step3: Calculate factorial values

We know that \(n!=n\times(n - 1)\times\cdots\times1\). So \(9!=9\times8\times7\times6\times5\times4\times3\times2\times1 = 362880\), \(3!=3\times2\times1=6\) and \(2!=2\times1 = 2\). Then \(\frac{9!}{3!2!}=\frac{362880}{6\times2}=30240\).

Answer:

30240