Question

If the word (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in the dictionary, then the position of word SMALL is
(A). ${{46}^{th}}$
(B). ${{59}^{th}}$
(C). ${{52}^{nd}}$
(D). ${{58}^{th}}$

Answer 1

Hint: We know the order of the letters that are used while positioning the words in the dictionary. The order is A, then L, then M, and then S. First find the number of words starting with the letter A. We have to form 5 letter words and since the first letter is A, So we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is $4!$ . But we have the letter L repeated twice. So, we have to divide it by
$2!$ . Similarly, find the numbers of words starting with the letter L, and M. Then find the number of words having the first two letters as SA and SL. The word “SMALL” is next to the position of the last word of the word having SL as their first two words. So, find the position of the last word having SL as its first two letters. When we have n letters and we have to find the number of possible ways to make words, then the number of possible words is $n!$ .

Complete step-by-step solution -
According to the question, it is given that the word we have is “SMALL”. In this word, we have the letters S, M, A, L, and L.
We know the order of the letters that are used while positioning the words in the dictionary. The order is A, then L, then M, and then S.
In the beginning, we have all the words starting with the letter A.
We have to form 5 letter words and since the first letter is A, So we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is $4!$ .
The number of words starting with the letter A = $4!$ .
But we have the letter L which is repeated twice here.
So, the number of words starting with the letter A = $\dfrac{4!}{2!}=\dfrac{4\times 3\times 2\times 1}{1\times 2}=12$ .
After all the words starting with letter A, we have the words starting with the letter L.
We have to form 5 letter words and since the first letter is L, So we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is $4!$ .
The number of words starting with the letter L = $4!=4\times 3\times 2\times 1=24$ .
After all the words starting with letter L, we have the words starting with the letter M.
We have to form 5 letter words and since the first letter is M, So we have 4 letters to be filled at the four positions. The number of ways that the four letters can be arranged is $4!$ .
The number of words starting with the letter M = $4!$ .
But we have the letter L which is repeated twice here.
The number of words starting with the letter M = $\dfrac{4!}{2!}=\dfrac{4\times 3\times 2\times 1}{1\times 2}=12$ .
After all the words starting with the letter M, we have the words having SA as their first two letters.
We have to form 5 letter words and since the first two letters of the words are SA, So we have 3 letters to be filled at the three positions. The number of ways that the three letters can be arranged is $3!$ .
But we have the letter L which is repeated twice here.
The number of words having SA as their first two letters = $\dfrac{3!}{2!}=\dfrac{3\times 2\times 1}{1\times 2}=3$ .
After the words having SA as their first two letters, we have the words whose first two letters are starting with the letter SL.
We have to form 5 letter words and since the first two letters of the words are SL, So we have 3 letters to be filled at the three positions. The number of ways that the three letters can be arranged is $3!$ .
The number of words having SL as their first two letters = $3!=3\times 2\times 1=6$ .
The position of the last word having SL as their first two letters = 12+24+12+3+6=57.
Now, we have the word “SMALL” as our next word.
The position of the word SMALL=57+1=58.
Therefore, the position of the word “SMALL” is ${{58}^{th}}$ .
Hence, the correct option is (D).

Note: In this question, one might forget to divide the number of words by $2!$ while finding the number of words starting with the letter L. When we have n letters and we have to find the number of possible ways to make words, then the number of possible words are $n!$ . But if we have m number of the same letters then the number of possible words will be $\dfrac{n!}{m!}$ .