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**Hint:**In this particular question use the concept that the number of ways to arrange n different object is equal to n!, and use the concept that if we want some letters not together so first find out all the words with these letters together then subtract these words from the total possible words, so use these concepts to reach the solution of the question.

__Complete step-by-step answer__:Given word:

‘NATIONAL’

As we see that all the letters in the given word are different except 1 and the number of letters are 8.

As we all know there are 5 vowels present in the English alphabets which are given as A, E, I, O and U.

Out of these alphabets the number of alphabets are present in the given word NATIONAL are A, I and O and A, as A is repeated two times

So in the given word there are 4 vowels and 4 consonants.

Now the total number of words which are possible form the letters of a given word NATIONAL = $\dfrac{{8!}}{{2!}}$, as A is repeated one time so divide by 2!

Consider that all the vowels are together so consider four vowels present in the given word as one letter.

So the arrangements of vowels internally = $\dfrac{{4!}}{{2!}}$

So there are 5 letters in the word so the number of arrangements = 5!

So the total number of words possible when all the vowels are together = \[\left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]

So the total number of words such that no vowel is together is the difference of total number of words from the given word and the total number of words when all the vowels are together.

Therefore, the total number of words such that no vowel is together = \[\dfrac{{8!}}{{2!}} - \left( {5! \times \dfrac{{4!}}{{2!}}} \right)\]

Now simplify we have,

\[ \Rightarrow \dfrac{{8.7.6.5!}}{{2.1}} - \left( {5! \times \dfrac{{4.3.2.1}}{{2.1}}} \right)\]

\[ \Rightarrow 4.7.6.5! - \left( {12.5!} \right)\]

\[ \Rightarrow 5!\left( {168 - 12} \right) = 120\left( {156} \right) = 18720\]

**Hence option (a) is the correct answer.**

**Note**: Whenever we face such types of questions the key concept we have to remember is that always recall that if there are n objects in which p objects are of same type and q objects are of another same type, so the number of ways to arrange them is given as $\dfrac{{n!}}{{p!\left( {q!} \right)}}$

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