Question

Position of the word ‘circle’ in dictionary formed by the words using the letters of the word ‘circle’ is:
A. $66$
B. $67$
C. $68$
D. None of these

Answer 1

In the given question, we are required to find out the rank of the word ‘CIRCLE’ in the dictionary of the words formed by the letters of the same word. So, we find the number of arrangements of the word ‘CIRCLE’ with or without meaning. The given question revolves around the concepts of permutations and combinations. We must have an idea about the multiplications rule of counting so as to solve such questions with ease. Then, we arrange all these words in the alphabetical order as in a dictionary and find the rank of the word. So, we list the letters of the given word in alphabetical order and then arrange them.

Complete step by step answer:
So, we are required to find the position of the word ‘CIRCLE’ in a dictionary formed by the words using the letters of the word ‘CIRCLE’. The number of letters in the word ‘CIRCLE’ is $6$. Now, there are two C’s in the word ‘CIRCLE’. So, we have to keep this in mind while arranging the letters of the word ‘CIRCLE’ so as to find the number of words. We know that the number of ways of arranging n things out of which r things are alike is $\left( {\dfrac{{n!}}{{r!}}} \right)$. So, the number of total arrangements of the letters of the word ‘CIRCLE’ is $\dfrac{{6!}}{{2!}} = \dfrac{{720}}{2} = 360$.
Hence, the number of words which can be made with the letters of the word ‘CIRCLE’ is $360$.

Now, we arrange all these $360$ words in alphabetical order as in a dictionary. The letters of the word ‘CIRCLE’ arranged in alphabetical order are C, C, E, I, L, R. Now, we start fixing the first two letters in the word in alphabetical order.If we fix the two C’s in the first two places of the word, we are left with $4$ places which are to be filled with E, I, L and R. So, the number of ways of filling the $4$ places with E, I, L and R is $4! = 24$. Hence, the number of words starting with two C’s is $24$.Similarly, if we fill up the first two places of the word with C and E, then the remaining $4$ places are to be filled with C, I, L and R. So, the number of words starting with C and E is $4! = 24$.

Now, we know that ‘CIRCLE’ starts with ‘CI’. So, we start fixing three letters in the starting of the word. If we fix the C, I, and C in the first three places of the word, we are left with $3$ places which are to be filled with E, L and R. So, the number of ways of filling the $3$ places with E, L and R is $3! = 6$. Similarly, if we fill up the first three places of the word with C, I and E, then the remaining $3$ places are to be filled with C, L and R. So, the number of ways of filling the $3$ places with C, L and R is $3! = 6$. Similarly, if we fill up the first three places of the word with C, I and L, then the remaining $3$ places are to be filled with C, E and R. So, the number of ways of filling the $3$ places with C, E and R is $3! = 6$.

Now, we know that ‘CIRCLE’ starts with ‘CIR’. So, we start fixing four letters in the starting of the word. So, we count the number of words till this. So, we get,
$\Rightarrow 24 + 24 + 6 + 6 + 6$
Adding up the numbers, we get,
$\Rightarrow 48 + 18$
$\Rightarrow 66$
Now, if we fix the C, I, R and C in the first four places of the word, we are left with $2$ places which are to be filled with E and L. So, we know that E comes before L in the alphabetical order. So, the ${67^{th}}$ word in the dictionary order of the words formed by the letters of CIRCLE is CIRCLE. So, we count the number of words before CIRCLE to find its position. Hence, the position of CIRCLE in a dictionary formed only using letters appearing in the word ‘CIRCLE’ is $68$.

Hence, option C is the correct answer.

Note: One should know about the number of arrangements in a word with a known number of letters. Care should be taken while handling the calculations. Calculations should be verified once so as to be sure of the answer. One must know that the number of ways of arranging n things out of which r things are alike is $\left( {\dfrac{{n!}}{{r!}}} \right)$. We must know how the words are arranged in a dictionary in order to solve these types of problems.