Answer

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**Hint:**We have to rephrase the condition of no vowel being in between two consonants as all the consonants being together. We take them as one input and try to arrange the rest of the alphabets. Finally, we arrange the consonants among themselves to find the solution to the problem.

**Complete step-by-step solution**

We have to find the total number of words that can be made by writing the letters of the word PARAMETER. But the condition is that no vowel is in between two consonants.

So, all the consonants have to be together. The vowels can be anywhere but not in between consonants.

The word PARAMETER has 5 consonants with 2 of them being similar and 4 vowels with 2 vowels having a copy of each.

Now to keep the consonants together we treat all 5 consonants as a single input.

So, instead of having 9 alphabets, we have now 5 of 4 vowels and 1 input.

We try to find the number of arrangements we can get from those 5 things out of which 4 are similar to 2 at a time.

The number of arrangements of n things of which r of a kind and k are of a kind is \[\dfrac{n!}{r!\times k!}\].

For our case we get it will be \[\dfrac{5!}{2!\times 2!}=\dfrac{120}{4}=30\].

Now the input of all the consonants also can arrange themselves among them.

There are 5 consonants with 2 being one of a kind.

So, their permutation will be \[\dfrac{5!}{2!}=\dfrac{120}{2}=60\].

**So, the total number of arrangements will be $60\times 30=1800$. The correct option is A.**

**Note:**The condition of no vowel being in between two consonants works for any number of vowels. So, all the consonants have to be together. We are taking the multiplication of both arrangements $60\times 30=1800$ as they are independent of each other.

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