Question

how many different ways can the letters of \personnel\ be arranged? if the letters of \personnel\ are arranged in a random order, what is the probability that the result will be \personnel\? the number of different ways that the letters of \personnel\ can be arranged is (type an integer or a simplified fraction )

Explanation:

Step1: Count total letters and repeated letters

The word "personnel" has 9 letters. The letter 'n' appears 2 times and the letter 'e' 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 \(n = 9\), \(n_1=2\) (for 'n') and \(n_2 = 2\) (for 'e'). So the number of arrangements is \(\frac{9!}{2!2!}\).
\[

$$\begin{align*} 9!&=9\times8\times7\times6\times5\times4\times3\times2\times1=362880\\ 2!&=2\times 1 = 2 \end{align*}$$

\]
\(\frac{9!}{2!2!}=\frac{362880}{2\times2}=90720\)

Step3: Calculate probability

There is only 1 correct arrangement (the word "personnel" itself). The probability \(P\) of getting the word "personnel" when arranging the letters randomly is the number of favorable outcomes (1) divided by the number of total outcomes (\(\frac{9!}{2!2!}\)). So \(P=\frac{1}{\frac{9!}{2!2!}}=\frac{2!2!}{9!}=\frac{4}{362880}=\frac{1}{90720}\)

Answer:

The number of different ways that the letters of "personnel" can be arranged is \(90720\).
The probability that the result will be "personnel" when arranged in a random order is \(\frac{1}{90720}\)