Word Formation with Repetition

Word Formation with Repetition (Permutation of Identical Objects) deals with arranging objects where some are identical. The number of distinct arrangements of 'n' objects where there are p identical objects of one type, q of another, etc., is n!/(p! × q! × ...). This is commonly applied to finding distinct anagrams of words with repeated letters.

10Worksheets

200+Practice Questions

Beginner to IntermediateDifficulty

2-3 hoursHours to Master

What You'll Learn

Introduction & Overview Why This Matters How to Solve Pro Tips & Tricks

Shortcut Methods Common Mistakes Practice Worksheets Exam Importance

Introduction to Word Formation with Repetition

Prerequisites

Basic permutation Factorial concept Understanding of identical objects Division of factorials

Why This Matters: Word formation problems appear in 1-2 questions in SSC CGL and Banking exams. They test handling of identical objects in permutations.

How to Solve Word Formation with Repetition Problems

Step 1: Count total number of letters/objects (n)

Step 2: Count frequency of each repeated letter/object

Step 3: Apply formula: n! / (p! × q! × r! × ...) where p, q, r are frequencies of repeats

Step 4: Calculate numerator and denominator carefully

Step 5: Cancel common factors to simplify

Step 6: For words with vowels/consonants constraints, handle restrictions first

Step 7: Verify that the answer is an integer

Pro Strategy: Always list all letter frequencies first. The denominator is the product of factorials of each repeated letter's frequency. For letters that appear once, their factorial (1! = 1) doesn't affect the calculation.

Example Problem

Example: How many distinct arrangements can be made from the letters of the word 'MISSISSIPPI'? Solution: Step 1: Total letters n = 11 Step 2: Letter frequencies: M=1, I=4, S=4, P=2 Step 3: Formula: 11! / (4! × 4! × 2! × 1!) Step 4: = 39916800 / (24 × 24 × 2) = 39916800 / 1152 = 34650 Answer: 34,650 distinct arrangements

Pro Tips & Tricks

For a word with all distinct letters, number of arrangements = n!
The denominator accounts for overcounting due to identical letters
If a letter repeats p times, divide by p!
For multiple repeated letters, divide by product of their factorials
Use cancellation to avoid large number calculations
The formula works for any objects with repetition, not just letters

Shortcut Methods to Solve Faster

For words with two repeated letters of frequencies a and b: n!/(a! × b!)

For words like 'BOOK' (O repeats twice): 4!/2! = 12

For words like 'BANANA': 6!/(3! × 2!) = 60

When calculating, cancel the larger factorial with part of the numerator

Common Mistakes to Avoid

Forgetting to divide by factorials of repeated letters

Including factorials for letters that appear once

Not counting the total letters correctly

Confusing permutations with repetitions (where repetition is allowed) vs permutations of identical objects

Practice Worksheets

Practice makes perfect! Work through these worksheets to master Word Formation with Repetition. Each worksheet contains 20 questions with detailed explanations. Start from Worksheet 1 and progress through increasing difficulty levels.

Worksheet 1 Beginner Beginner level - Perfect for building foundational understanding

Worksheet 2 Beginner Beginner level - Perfect for building foundational understanding

Worksheet 3 Beginner Beginner level - Perfect for building foundational understanding

Worksheet 4 Intermediate Intermediate level - Test your understanding with moderate difficulty

Worksheet 5 Intermediate Intermediate level - Test your understanding with moderate difficulty

Worksheet 6 Intermediate Intermediate level - Test your understanding with moderate difficulty

Worksheet 7 Advanced Advanced level - Challenge yourself with complex problems

Worksheet 8 Advanced Advanced level - Challenge yourself with complex problems

Worksheet 9 Advanced Advanced level - Challenge yourself with complex problems

Worksheet 10 Advanced Advanced level - Challenge yourself with complex problems