Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Concept 🎬 Subsets with No Consecutive Numbers — JEE in 60 Sec Set ke subsets banana hai… lekin ek strict rule — 👉 koi do numbers consecutive nahi hone chahiye! JEE mein yeh pattern baar-baar repeat hota hai, aur iska answer hamesha Fibonacci logic se aata hai 🔥 🖼️ Concept Image ✍️ Short Explanation Is concept ka core idea bahut simple hai: 👉 har number ke paas sirf 2 choices hoti hain — include ya exclude. Bas ek restriction add ho jaata hai: include karte hi next number block ho jaata hai . 🔹 Step 1 — Problem Type Given set: { 1 , 2 , 3 , … , n } Allowed: Any subset But NO two consecutive elements Examples: { 1 , 3 , 5 } \{1,3,5\}  ✔ allowed { 1 , 2 , 4 } \{1,2,4\}  ❌ not allowed (1 & 2 consecutive) 🔹 Step 2 — Define a Counting Function Let: f ( n ) = number of valid subsets of  { 1 , 2 , … , n } 📌 Ab poora question f(n) nikaalne ka hai. 🔹 Step 3 — Key Idea (MOST IMPORTANT 🔥)** Element ...

  ❓ Question For n ≥ 2 n \ge 2 , let S n S_n ​ denote the set of all subsets of { 1 , 2 , … , n } \{1,2,\dots,n\} with no two consecutive numbers . For example: { 1 , 3 , 5 } ∈ S 6 , { 1 , 2 , 4 } ∉ S 6 . \{1,3,5\}\in S_6,\quad \{1,2,4\}\notin S_6. Then the value of n ( S 5 ) n(S_5) (i.e. the number of elements in S 5 S_5 ​ ) is equal to ? 🖼️ Question Image ✍️ Short Solution This is a classic JEE combinatorics + recurrence question. Key idea: 👉 Subsets with no consecutive integers follow the Fibonacci pattern. We’ll first understand why , then apply it to n = 5 n=5 . 🔹 Step 1 — Understand the restriction We are forming subsets of { 1 , 2 , 3 , 4 , 5 } \{1,2,3,4,5\}  such that: You may include a number But if you include k k , you cannot include k + 1 k+1 So every choice blocks the next number. 🔹 Step 2 — Define a counting approach Let: f ( n ) = number of subsets of  { 1 , 2 , … , n }  with no consecutive...