How do you find the rank of a word in a dictionary?

Step 2: Find out the number of letters on the right with lower rank  As G is 3 letter in alphabetical order and there are 2 letters on the right side with lower rand than G so we write 2 as its value and A is ranked 1 and there is no letter on right side of A which is lower rank than A so value is 0 

Step 3: Write 0! from right end and continue till left end with increasing order 

Such as we start with 0! and reached 5! for first value.

Step 4: Multiply with respective value and get the answer.

240 + 0 + 12 + 0 + 2 + 0 =254

Hint: To solve this problem, we will use the fact that dictionary format uses the alphabetical format. Thus, we will try to find the number of combinations of the words that can be made from the letters M, O, T, H, E and R which are in alphabetical order below the word, ‘MOTHER’. This will help us provide the rank of the word ‘MOTHER’.

Complete step-by-step answer:
For solving this question, we will first try to find the total number of combinations that can be made from the letters M, O, T, H, E and R. In general, the formula for this is n! (in this case, it is 6! = 720). However, this is clearly not the rank of the word ‘MOTHER’ since this would not be the last word alphabetically made from the letters M, O, T, H, E and R (in fact, the last word would be TROMHE). So, we start with finding the list of letters alphabetically below, ‘MOTHER’. We start with the first lowest letter which is E. Thus, the letters made with E as the first letter would be 5! = 120 (since, first place, E is fixed and thus only the next five letters change the permutation).
Similarly, the next lowest letter after E is H. Thus, similarly, if we fix H at first place, again the number of combinations would be 5! = 120. After this, the next letter after H is M. Now, in this case, this aligns with the required word (MOTHER).
Now, instead of calculating all the combinations. We first fix the second letter. Alphabetically, this would be E (Now, the first two letters that are ME are fixed). Now, the total number of combinations with these two letters fixed are 4! = 24.
Next, we fix the second letter alphabetically (which would be H, thus now MH is fixed). Again the combinations would be 4! = 24.
Alphabetically, the next available letter would be O (since, M is already occupied in first place). Again, MO aligns with the word ‘MOTHER’. So, now, we will try to fix 3 letters and then find the combinations. Again, alphabetically, we have E (thus, fixing MOE), we have total combinations as 3! = 6.
Next, the letter alphabetically would be H (thus, fixing MOH), we have total combinations as 3! = 6.
Next, available letters alphabetically would be R (thus, fixing MOR), we have total combinations as 3! = 6.
Now, the next letter would be T. This aligns with the word ‘MOTHER’. Now, we fix 4 letters. Thus, the first available letter would be E (MOTE is fixed). Thus, we have 2 combinations possible (2! = 2). Now, the next letter would be H, thus, again MOTH aligns with the word MOTHER. Now, the next available letter would be E (thus, MOTHE is fixed). However, this again aligns with the word MOTHER. Thus, we have all the combinations.
Thus, we first sum up all the total number of combinations we have. This is given by 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 = 308. Since, there are 308 combinations before the word, ‘MOTHER’. The rank of the word MOTHER is 308 + 1 =309.

Note: Generally, solving problems involving dictionaries is similar to having a set of sorted words (in terms of alphabets). The only task we have to do is to then find at which position the word lies on these lists of sorted words. For this, in manual calculations (that is by permutations and combination), we have to find all the number of combinations before the required word. In case of non-manual methods (such as in computer algorithms containing the list of these sorted words), we can use methods like binary search to find the rank of the word more efficiently and quicker than the method we used in the solution above.

Now in the dictionary words will appear in alphabetical order, so the first words will appear starting alphabet “A“.

When A is fixed at the first position, the rest 4 alphabets can be arranged in 4! = 24 ways.

The next starting alphabet will be “M” and again there will be 4! = 24 words starting with “M“.

The next starting alphabet will be “R” and again there will be 4! = 24 words starting with “R“.

Next will be starting with “S“, and next alphabet as “M” and next as “A” will have 2! = 42 words but the first word will be S > M > A > R > T which is the desired word.

For some students "Rank finding in dictionary problems" is a difficult topic in quantitative aptitude. Actually it is not like that. Once we understand the concept, it is not a difficult one to understand. 

How to Find the Rank of a Word in Dictionary With Repetition - Practice question

Question 1 :

Find the number of strings that can be made using all letters of the word THING. If these words are written as in a dictionary, what will be the 85th string?

Solution :

The given string "THING" has 5 letters, there is no repetition of letters.

So, the number of strings can be made using the above 5 letters,

T, H, I, N, G   =  5!  =  120.

Now we have to find the word in the 85th place. For that let us count the words that can be formed using the letters

Alphabetical order of the word  G, H, I, N, T

Number of words formed starting with "G"

G  __ __ __ __  =  4!  =  24

Number of words formed starting with "H"

H  __ __ __ __  =  4!  =  24

Number of words formed starting with "I"

I  __ __ __ __  =  4!  =  24

So far, we get 72 words. Hence the required word starts with the letter N.

Number of words starting with the letters "NG"

N  G  __  __  __  =  3!  =  6

Number of words starting with the letters "NI"

N  I  __  __  __  =  3!  =  6

So far, we get 84 words. 

By writing the letters after N I in alphabetical order, we get the word "NIGHT".

Hence it must the required word at the 85th place.

Question 2 :

If the letters of the word FUNNY are permuted in all possible ways and the strings thus formed are arranged in the dictionary order, find the rank of the word FUNNY.

Solution : 

Alphabetical order of the word "FUNNY" 

F, N, N, U, Y

Number of words starting with "FN" 

F  N  __ __ __  =  3!  =  6

Number of words starting with "FUN". By arranging the remaining letters in the alphabetical order, we get 

F U N N Y  =  1

Hence the rank of the word "FUNNY" is 7.

After having gone through the stuff given above, we hope that the students would have understood "How to Find the Rank of a Word in Dictionary With Repetition". 

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