
How many 4-letter words, with or without meaning, can be formed out of the letters of the word, “LOGARITHM”, if repetition of letter is not allowed.
Answer
493.8k+ views
Hint: This is a question based on permutations and combinations. If you see all the alphabets in the word LOGARITHM are different, so we have 9 alphabets. So, actually we need to find the number of permutations of 4 out of these 9 alphabets using the formula $^{9}{{P}_{4}}$ .
Complete step-by-step solution -
Let us start by interpreting the things given in the question. The word given to us is “LOGARITHM”. It consists of L, O, G, A, R, I, T, H, M, i.e., 9 different alphabets and we are asked to find the number of 4-letter words that can be formed using these 9 alphabets. So, basically we need to find the number of permutations of 4 alphabets out of 9 alphabets possible.
We know that the permutation of r different objects out of n different objects is given by $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .
Therefore, for the number of permutations of 4 different alphabets, out of the 9 different alphabets of the word LOGARITHM, n=9 and r=4. The number of such permutations is given by $^{n}{{P}_{r}}=\dfrac{9!}{\left( 9-4 \right)!}=\dfrac{9!}{5!}=9\times 8\times 7\times 6=3024$ .
Note: In questions related to permutations and combinations, students generally face problems in deciding whether the question is based on selection or arrangement. Also, be careful if the repetition of letters is allowed or not, as the answers to both the cases are different. For instance: if the repetition of alphabets was allowed in the above question, the answer would be $9\times 9\times 9\times 9=6561$ .
Complete step-by-step solution -
Let us start by interpreting the things given in the question. The word given to us is “LOGARITHM”. It consists of L, O, G, A, R, I, T, H, M, i.e., 9 different alphabets and we are asked to find the number of 4-letter words that can be formed using these 9 alphabets. So, basically we need to find the number of permutations of 4 alphabets out of 9 alphabets possible.
We know that the permutation of r different objects out of n different objects is given by $^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$ .
Therefore, for the number of permutations of 4 different alphabets, out of the 9 different alphabets of the word LOGARITHM, n=9 and r=4. The number of such permutations is given by $^{n}{{P}_{r}}=\dfrac{9!}{\left( 9-4 \right)!}=\dfrac{9!}{5!}=9\times 8\times 7\times 6=3024$ .
Note: In questions related to permutations and combinations, students generally face problems in deciding whether the question is based on selection or arrangement. Also, be careful if the repetition of letters is allowed or not, as the answers to both the cases are different. For instance: if the repetition of alphabets was allowed in the above question, the answer would be $9\times 9\times 9\times 9=6561$ .
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