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**Hint:**We first discuss the form by which we can take 4 letters from ‘LOGARITHMS’ as all of them are unique letters. Then we place the values for $n=10;r=4$ in ${}^{n}{{P}_{r}}$. We complete the multiplication and find the solution.

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

We have the main word ‘LOGARITHMS’. All of its letters are unique. We make 4 letter words, with or without meaning. We first take 4 letters out of those 10 letters and arrange them.

The given mathematical expression ${}^{n}{{P}_{r}}$ is an expression of permutation.

We replace the values $n=10;r=4$ in ${}^{n}{{P}_{r}}$.

The simplified form of the mathematical expression ${}^{n}{{P}_{r}}$ is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.

Here the term $n!$ defines the notion of multiplication of first n natural numbers.

This means $n!=1\times 2\times 3\times ....\times n$.

The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation.

Now we try to find the value of ${}^{10}{{P}_{4}}$. We put the values of $n=10;r=4$ and get ${}^{10}{{P}_{4}}=\dfrac{10!}{\left( 10-4 \right)!}$.

We now solve the factorial values.

\[{}^{10}{{P}_{4}}=\dfrac{10!}{6!}=\dfrac{10\times 9\times 8\times 7\times 6!}{6!}=5040\].

Therefore, the number of 4-letter words, with or without meaning is 5040.

The correct option is (a).

**Note:**There are some constraints in the form of ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the combination happens first even though we are finding permutation. The choosing of the $r$ objects happens first, then we arrange them. The mathematical expression is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\times \left( n-r \right)!}\times r!={}^{n}{{C}_{r}}\times r!$.

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