Question

Find the rank of the word ‘MOTHER’ in dictionary format.

Answer 1

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.