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**Hint:**Here, the given question is from permutation and combination. We have to find the number of arrangements that can be formed out by the letters of the word “LOGARITHM” considering different conditions. So, firstly we have to count the number of consonants and vowels in the word “LOGARITHM”, then apply the conditions such as in the first question arrange the consonant by putting a space between two consonants and then arrange the vowels in that place.

**Complete step-by-step solution:**

(1) Here, the letters of the word “LOGARITHM” are arranged in such a way that no two vowels come together.

Since, the given condition is that no two vowels come together and total six consonants and three vowels are in the word “LOGARITHM” so the possible arrangement may be VCVCVCVCVCVCV. And of the six positions of consonant, six consonants can be arranged in $6!$ ways. The arrangement of six consonants produces seven places for vowels and we have only three vowels so this can be arranged as $7{p_3}$.

So, the total possible number of arrangements is $6!\,7{p_3}$.

**Hence, option (A) is correct.**

(2) Here, the letters of the word “LOGARITHM” are arranged in such a way that all the vowels do not come together.

To arrange according to given condition, find the total possible number of arrangements of letters without any condition that is $9!$ and then subtract the case when all the vowels come together from this.

Now, put all vowels in to a box and consider it as a single letter and make arrangements of six consonants and a box of vowels that comes as $7!\,3!$ because the seven letters can be arranged in $7!$ and the vowels inside the box can be arranged internally in $3!$ ways.

So, the total possible number of arrangements is $9! - 7!\,3!$

**Hence, option (B) is correct.**

(3) Here, the letters of the word “LOGARITHM” are arranged in such a way that no two consonants come together.

We have only three vowels but six consonants and we have to arrange such that no two consonants come together and this can be possible if we arrange the letters as CVCVCVC but there is only four places for consonants but we have to arrange six consonants so, it is not possible to arrange in such a way.

So, the possible number of arrangements is zero.

**Hence, option (A) is correct.**

**Note:**We should know,

$n{p_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\,}}$

\[n{C_r} = \dfrac{{n!}}{{\left( {n - r} \right)!\,r!}}\]

It will be helpful in finding the numerical value.

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