Question

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

Explanation:

Step1: Count total letters and repeated letters

The word "difference" has 9 letters. The letter 'f' appears 2 times and the letter 'e' appears 3 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 each repeated object is \(\frac{n!}{n_1!n_2!\cdots n_k!}\). Here \(n = 9\), \(n_1=2\) (for 'f') and \(n_2 = 3\) (for 'e'). So the number of arrangements is \(\frac{9!}{2!×3!}\).

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\(\frac{9!}{2!×3!}=\frac{362880}{2\times6}=30240\).

Step3: Calculate probability of alphabetical order

There is only 1 way for the letters to be in alphabetical order out of the 30240 possible arrangements. So the probability is \(\frac{1}{30240}\).

Answer:

The number of different ways that the letters of "difference" can be arranged is 30240.
The probability that the letters will be in alphabetical order is \(\frac{1}{30240}\).