In this video we solve Question 10 of Exercise 6.3 from Class 11 Maths — Chapter 6 (Permutations & Combinations, NCERT 2025–26) step by step. 🔖 Question: In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together? 🧩 Here’s the algorithm you’ll learn step-by-step: 1️⃣ Identify total letters and find the total distinct permutations (because of repeated letters). 2️⃣ Group all 4 I’s together and treat them as one unit to count cases where they come together. 3️⃣ Use the difference method → (Total – Together = Not Together) to calculate the required number of arrangements. 4️⃣ Understand how the concept of restrictions changes the counting logic in permutation problems. 🧮 Method used (clear steps): 👉 1. Find total distinct permutations of MISSISSIPPI using the formula for permutations with repetition: Total = 11!/4!4!2!=34,650 👉 2. Count permutations where the four I’s are together (treat the 4 I’s as a single block). Then arrange this block + remaining letters (S repeated 4, P repeated 2): 4 I’s together =8!/4!2! = 840 👉 3. Subtract to get permutations where the four 4 I’s do not come together: 34,650−840=33,810 🎯 Final Answer: 33,810 permutations where the four I’s are not together. 💡Why This Video Is Important: Demonstrates how to handle repeated letters in permutations (a common board/competition pattern). Shows the block method vs. total-minus-block approach — a fast, reliable technique for many permutation problems. Time-saving for Class 11 Board Exams, JEE Main, and CUET — you’ll learn a method that avoids long casework and reduces mistakes. ✅ What you’ll learn by watching: Use of factorials with repeated items. When and how to use the block method. How to combine both methods (total minus block) to get the required count quickly and safely. #Class11Maths #Exercise6_3 #PermutationsAndCombinations #MISSISSIPPI #WordPermutations #PermutationWithRepetition #BlockMethod #NCERT2025 #JEEMainPrep #CUETPrep #MathsTricks #YourClassroom