Rank of a Word in a Dictionary: A Systematic Approach
Master the step-by-step algorithm to find dictionary rank of words when all letters are arranged alphabetically.
Understanding Dictionary Rank Problems
Dictionary rank problems ask: "If all permutations of letters of a word are arranged in alphabetical order, what is the position (rank) of the given word?"
This is a classic JEE permutation problem that tests your understanding of:
- Factorial concepts and permutations
- Systematic counting approaches
- Alphabetical ordering principles
- Step-by-step problem solving methodology
The 4-Step Algorithm
Step 1: Arrange Letters Alphabetically
Write all letters of the word in alphabetical order. This gives you the reference for comparison.
Step 2: Fix Position & Count Smaller Words
For each position in your word, count how many words would come before it by considering letters smaller than the current letter that haven't been used yet.
Formula: (Number of smaller unused letters) × (Remaining letters factorial)
Step 3: Move to Next Position & Repeat
Move to the next letter, mark the current letter as used, and repeat Step 2 until you reach the end of the word.
Step 4: Add 1 to Get Final Rank
Add all the counts from previous steps and add 1 (since rank starts from 1, not 0).
Example 1: Rank of "QUESTION"
Find the dictionary rank of the word "QUESTION" when all permutations of its letters are arranged alphabetically.
Step-by-Step Solution:
Step 1: Alphabetical order: E, I, N, O, Q, S, T, U
Step 2: Fix positions and count:
• Position 1 (Q): Letters smaller than Q: E, I, N, O (4 letters)
Count = 4 × 7! = 4 × 5040 = 20160
Step 3: Move to position 2 (U): Used letters: Q, Remaining: E, I, N, O, S, T, U
Letters smaller than U: E, I, N, O, S, T (6 letters)
Count = 6 × 6! = 6 × 720 = 4320
Step 4: Continue this process...
Position 3 (E): No smaller letters → Count = 0
Position 4 (S): Smaller than S: I, N, O (3 letters)
Count = 3 × 4! = 3 × 24 = 72
Position 5 (T): Smaller than T: I, N, O (3 letters)
Count = 3 × 3! = 3 × 6 = 18
Position 6 (I): No smaller letters → Count = 0
Position 7 (O): Smaller than O: N (1 letter)
Count = 1 × 1! = 1
Position 8 (N): Last position → Count = 0
Final Calculation:
Total words before = 20160 + 4320 + 0 + 72 + 18 + 0 + 1 + 0 = 24571
Rank = 24571 + 1 = 24572
Example 2: Rank of "SURYA"
Find the dictionary rank of the word "SURYA" when all permutations are arranged alphabetically.
Step-by-Step Solution:
Step 1: Alphabetical order: A, R, S, U, Y
Step 2: Position 1 (S): Letters smaller than S: A, R (2 letters)
Count = 2 × 4! = 2 × 24 = 48
Step 3: Position 2 (U): Used: S, Remaining: A, R, U, Y
Letters smaller than U: A, R (2 letters)
Count = 2 × 3! = 2 × 6 = 12
Step 4: Position 3 (R): Used: S, U, Remaining: A, R, Y
Letters smaller than R: A (1 letter)
Count = 1 × 2! = 1 × 2 = 2
Step 5: Position 4 (Y): Used: S, U, R, Remaining: A, Y
Letters smaller than Y: A (1 letter)
Count = 1 × 1! = 1
Final Calculation:
Total words before = 48 + 12 + 2 + 1 = 63
Rank = 63 + 1 = 64
🚀 Quick Solving Strategies
For Words with Repeated Letters:
- Divide by factorial of repetitions
- Example: "MISSISSIPPI" has 4 I's, 4 S's, 2 P's
- Total permutations = 11! ÷ (4! × 4! × 2!)
- Apply same algorithm but account for duplicates
Common Mistakes to Avoid:
- Forgetting to add 1 at the end
- Not marking used letters properly
- Missing smaller letters in counting
- Calculation errors in factorials
Examples 3-4 Available in Full Version
Includes "BANANA" with repeated letters and "MATHEMATICS" with advanced techniques
📝 Quick Self-Test
Try these JEE-level problems to test your understanding:
1. Find rank of "APPLE"
2. Find rank of "GOOGLE" (with repeated letters)
3. Find rank of "INDIA"
Answers: 1. 44, 2. 88, 3. 42
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