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 \(n_1,n_2,\cdots,n_k\) are the number of times of repeated objects 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 \(\frac{9!}{2!2!}\).

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

Step3: Calculate probability

There is only 1 correct arrangement ("personnel"). The probability \(P\) of getting the correct arrangement when arranging the letters randomly is \(\frac{1}{90720}\).

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 \(\frac{1}{90720}\)