Answer

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**Hint:**

Here, we will find the number of ways using the concept of factorial. We will find the total Number of ways of arrangement of the letters. Then we will find the number of ways that two I’s come together. We will then subtract the number of ways that two I’s come together from the total number of arrangements and simplify it further to get the required answer.

Formula Used:

We will use the following formulas:

1) Total Number of ways is given by formula \[N = \dfrac{{n!}}{{r!}}\] where \[n!\] are the total number of cases and \[r!\] are the likely events.

2) Factorial representation: \[n! = n \times \left( {n - 1} \right)!\]

**Complete step by step solution:**

We are given letters of the word “UNIVERSITY”.

Now, we have to find the total number of ways of arrangement of letters.

Total number of letters in the word “UNIVERSITY” is 10.

The letter ‘I’ is repeated twice, so the number of likely events is 2.

Total Number of ways is found using the formula \[N = \dfrac{{n!}}{{r!}}\] .

\[ \Rightarrow \] Total Number of ways of arrangement of the letters \[ = \dfrac{{10!}}{{2!}}\]

Now, we have to find the number of ways that two I’s come together.

If two I’s come together, then it can be considered as a single identity, so the number of letters becomes 9.

Number of ways that two I’s come together \[ = 9!\]

Now, we have to find the number of ways that two I's do not come together by subtracting the number of ways that two I’s come together from the total number of ways of the arrangement of the letters.

Number of ways that two I’s that do not come together \[ = \] Total number of ways of arrangement of the letters \[ - \] Number of ways that two I’s come together

Substituting the values in the above equation, we will get

\[ \Rightarrow \] Number of ways that two I’s that do not come together \[ = \dfrac{{10!}}{{2!}} - 9!\]

By using Factorial representation, \[n! = n \times \left( {n - 1} \right)!\] , we get

\[ \Rightarrow \] Number of ways that two I’s that do not come together \[ = \dfrac{{10 \times 9!}}{2} - 9!\]

Taking out common factor and dividing, we get

\[ \Rightarrow \] Number of ways that two I’s that do not come together \[ = 9!\left( {\dfrac{{10}}{2} - 1} \right)\]

Dividing 10 by 2, we get

\[ \Rightarrow \] Number of ways that two I’s that do not come together \[ = 9!\left( {5 - 1} \right)\]

Subtracting the terms, we get

\[ \Rightarrow \] Number of ways that two I’s that do not come together \[ = 4 \times 9!\]

Therefore, the possible number of ways is \[4 \times \left| \!{\underline {\, 9 \,}} \right. \].

**Thus Option (D) is the correct answer.**

**Note:**

We know that factorial is defined as the multiplication of numbers in the descending order till unity. Factorial is also used in finding the number of ways of arrangement of letters, numbers, etc., as like permutations and combinations. Permutation is defined as the arrangement of something in an order where combinations are defined as selection of items where order is not important. Factorial can be represented in both the notations as \[\left| \!{\underline {\,a \,}} \right. \] and \[a!\].

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