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How many 4-letter words, with or without meaning, can be formed out of the letters of the word ‘LOGARITHMS’, if repetition of letters is not allowed?(a) 5040(b) 1000(c) 2500(d) 2060

Last updated date: 13th Jul 2024
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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 ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\times \left( n-r \right)!}\times r!={}^{n}{{C}_{r}}\times r!$.