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"Find the rank of the word MOTHER if all its permutations are arranged in dictionary order." It is a common question in the permutations chapter, and it is more about careful bookkeeping than clever ideas. Most wrong answers come from forgetting the "+1" at the end or from mishandling repeated letters. This post gives a method that works every time and checks it on two classic words.
Where does this problem come from?
The JEE (Main) 2026 syllabus lists the fundamental principle of counting, permutations and combinations, and "simple applications". NCERT Class 11 Chapter 6, Permutations and Combinations includes Example 22, which finds the 50th word in the dictionary list of arrangements of AGAIN, and Miscellaneous Exercise Question 4, which asks how many arrangements of EXAMINATION come before the first word starting with E. Check your exam year's syllabus. For the counting basics, see our guide to permutations and combinations.
What is the method?
Write the letters in alphabetical order.
Go through the word one position at a time.
At each position, for every unused letter that comes before the actual letter, count the arrangements of the remaining letters. Add them up.
Fix the actual letter and move to the next position.
Rank = total words before + 1.
Worked example 1: MOTHER (no repeated letters)
Alphabetical order: E, H, M, O, R, T.
First letter M. Smaller letters available: E, H. Each gives 5! = 120 words. Count so far: 240.
Second letter O, with E, H, O, R, T left. Smaller: E, H. Each gives 4! = 24. Count: 240 + 48 = 288.
Third letter T, with E, H, R, T left. Smaller: E, H, R. Each gives 3! = 6. Count: 288 + 18 = 306.
Fourth letter H, with E, H, R left. Smaller: E. That gives 2! = 2. Count: 308.
Fifth letter E, with E, R left. Nothing smaller. Count stays 308.
The last letter R is forced.
308 words come before MOTHER, so its rank is 309.
Worked example 2: SUCCESS (repeated letters)
Letters: C, C, E, S, S, S, U. With repeats, the number of arrangements of the remaining letters is the factorial of how many are left, divided by the factorial of each repeat count.
First letter S. Smaller letters available: C and E. Starting with C, the rest are C, E, S, S, S, U, giving 6!/3! = 120. Starting with E, the rest are C, C, S, S, S, U, giving 6!/(2! 3!) = 60. Count: 180.
Second letter U, with C, C, E, S, S, U left. Smaller distinct letters: C, E, S. SC leaves C, E, S, S, U, giving 5!/2! = 60. SE leaves C, C, S, S, U, giving 5!/(2! 2!) = 30. SS leaves C, C, E, S, U, giving 5!/2! = 60. Count: 180 + 150 = 330.
Third letter C and fourth letter C: nothing smaller is left. Fifth letter E, with E, S, S left: nothing smaller. The last two letters are S, S, one arrangement.
330 words come before SUCCESS, so its rank is 331.
The key detail: count each distinct smaller letter once, not each copy. Two Cs available still give one "starts with C" block.
The reverse problem: finding the nth word
NCERT's Example 22 runs the method backwards. AGAIN has 5!/2! = 60 arrangements. Words starting with A: 4! = 24. With G: 4!/2! = 12. With I: 12. That makes 48, so the 49th word is the first starting with N, NAAGI, and the 50th is NAAIG.
How can you check your answer?
Test the method on a short word you can list by hand. For CAT, the six arrangements in order are ACT, ATC, CAT, CTA, TAC, TCA, so CAT is third. The method gives: first letter C, smaller letter A gives 2! = 2, then nothing else, so rank 2 + 1 = 3. Agreement on small cases is good evidence that your bookkeeping is right. Eizenberg and Zaslavsky (2004) studied 14 undergraduates solving counting problems and describe verifying a solution as a major difficulty; of the 5 checking strategies they identified, 2 were more frequent and more helpful than the others.
Common mistakes and how to avoid them
Forgetting the +1.
Using all letters instead of unused ones at each step. Cross letters off as you fix them.
Counting a repeated letter twice as a starting option.
Dividing by the wrong factorials. Recount the repeats among the remaining letters at every step, since they change.
A practice routine
Find the ranks of three words a week: one with distinct letters, two with repeats.
Check each method on a three- or four-letter word you can list.
Solve one reverse problem, finding the nth word.
Try NCERT's EXAMINATION question: before the first word starting with E come all words starting with A, which is 10!/(2! 2!) = 907200.
When a step does not make sense, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Next step: find the rank of the word DAUGHTER in dictionary order, then check your answer of 5675 with the small-case test first.
