
How many four letter words with or without meaning, can be formed by using the letters of the word, 'LOGARITHMS', such that the letter R is always included ? ( repetition is not allowed)
A. $12096$
B. $2458$
C. $1296$
D. $2240$
Answer
520.2k+ views
Hint: In order to deal with this question, we will use the concept i.e., there are 9 letters to choose from and the first letter of the new word is fixed because R is fixed among 4 words, then 8 for the second, 7 for the third, and 4 words can be shuffled to each others, so that can be calculated by \[4!\].
Complete step-by-step answer:
Given word is LOGARITHMS
Number of letters in the given word ‘LOGARITHMS’ is 10
We basically have to form four letter words using the letters in the given word such that the letter R is always there.
In the given word ‘LOGARITHMS’, none of the letters is repeating (i.e., all are different alphabets). The letter ‘R’ needs to be there in the newly formed four digit word so out of 10 letters we are left with 9 letters from which we need to select any 3 letters.
Out of 9 letters (L,O,G,A,I,T,H,M,S), we can select any one as one letter of the four letter word which needs to be formed. This can be done by selecting any one letter out of these 9 letters and hence, 9 ways. We are left with 8 letters out of which any one can be selected in 8 ways. Finally, we are left with 7 letters out of which any one can be selected as the last letter of the four letter word in 7 ways.
So, the number of ways \[ = \;9 \times 8{\text{ }} \times {\text{ }}7 = {\text{ }}504\]
Now, these four letters can be shuffled within themselves and can be calculated by multiplying \[4{\text{ }}!\] with the total number of ways obtained above.
Hence, total number of ways to form $4$ letter new word \[ = {\text{ }}4!{\text{ }} \times {\text{ }}504{\text{ }} = {\text{ }}12096\]
So, the correct answer is “Option A”.
Note: In these types of problems, it is very important to know what a factorial means. In mathematics, the factorial of a positive integer n, denoted by \[n!,\] is the product of all positive integers less than or equal to n for example, The value of \[0!\] and \[1!\] is 1, according to the convention for an empty product.
Complete step-by-step answer:
Given word is LOGARITHMS
Number of letters in the given word ‘LOGARITHMS’ is 10
We basically have to form four letter words using the letters in the given word such that the letter R is always there.
In the given word ‘LOGARITHMS’, none of the letters is repeating (i.e., all are different alphabets). The letter ‘R’ needs to be there in the newly formed four digit word so out of 10 letters we are left with 9 letters from which we need to select any 3 letters.
Out of 9 letters (L,O,G,A,I,T,H,M,S), we can select any one as one letter of the four letter word which needs to be formed. This can be done by selecting any one letter out of these 9 letters and hence, 9 ways. We are left with 8 letters out of which any one can be selected in 8 ways. Finally, we are left with 7 letters out of which any one can be selected as the last letter of the four letter word in 7 ways.
So, the number of ways \[ = \;9 \times 8{\text{ }} \times {\text{ }}7 = {\text{ }}504\]
Now, these four letters can be shuffled within themselves and can be calculated by multiplying \[4{\text{ }}!\] with the total number of ways obtained above.
Hence, total number of ways to form $4$ letter new word \[ = {\text{ }}4!{\text{ }} \times {\text{ }}504{\text{ }} = {\text{ }}12096\]
So, the correct answer is “Option A”.
Note: In these types of problems, it is very important to know what a factorial means. In mathematics, the factorial of a positive integer n, denoted by \[n!,\] is the product of all positive integers less than or equal to n for example, The value of \[0!\] and \[1!\] is 1, according to the convention for an empty product.
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