1. Class 11
2. Important Question for exams Class 11
3. Chapter 7 Class 11 Permutations and Combinations

Transcript

Example 23 (Method 1) How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? There are total 7 digits in 1000000 We need to form a 7 digit number using the digits 1, 2, 0, 2, 4,2, 4 But, these include numbers starting with ‘0’ like 0412224, …etc which are actually 6 digit numbers Hence, we cant have number beginning with 0 So the number can begin either with 1, 2 or 4 Required numbers = Numbers starting with 1 + Numbers starting with 2 + Numbers starting with 4 Case 1 If number begin with 1 The remaining digit to be arranged will be 0, 2, 2, 2, 4, 4 Here three 2s & two 4s Since digit are repeating hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & two 4s & p1 = 3, p2 = 2 The numbers of numbers beginning with 1 = 𝑛!/𝑝1!𝑝2! = 6!/(3! 2!) = (6 × 5 × 4 × 3!)/(3! × 2 × 1) = 60 Case 2 If number begin with 2 The remaining digit to be arranged will be 1,0, 2, 2, 4, 4 Here two 2s & two 4s Since digit are repeating hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of remaining digit = 6 Thus, , n = 6 Since, two 2s & two 4s & p1 = 2, p2 = 2 The numbers of numbers beginning with 2 = 𝑛!/𝑝1!𝑝2! = 6!/(2! 2! ) = (6 × 5 × 4 × 3 × 2 × 1)/((2 × 1) × (2 × 1) ) = 180 Case 3 If number begin with 4 The remaining digit to be arranged will be 1,0, 2, 2, 2, 4 Here three 2s Since digit are repeating hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & p1 = 3 The numbers of numbers beginning with 4 = 𝑛!/(𝑝1! ) = 6!/3! = (6 × 5 × 4 × 3!)/3! = 120 Required numbers = Numbers starting with 1 + Numbers starting with 2 + Numbers starting with 4 = 60 + 180 +120 = 360 Example 23 (Method 2) How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4? There are total 7 digits in 1000000 We need to form a 7 digit number using the digits 1, 2, 0, 2, 4,2, 4 But, these include numbers starting with ‘0’ like 0412224, …etc which are actually 6 digit numbers Hence, we cant have number beginning with 0 Required numbers = All arrangements – Numbers starting with 0 All arrangements The digits to be arranged are 1, 0, 2, 2, 2, 4, 4 Here, three 2s & two 4s Since digit are repeating hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of digits = 7 Thus, , n = 7 Since, three 2s & two 4s & p1 = 3, p2 = 2 All arrangements = 𝑛!/𝑝1!𝑝2! = 7!/3!2! = (7 × 6 × 5 × 4 × 3!)/(3! × 2 × 1) = 420 If number begin with 0 The remaining digit to be arranged will be 1, 2, 2, 2, 4, 4 Here three 2s & two 4s Since digit are repeating hence we use this formula = 𝑛!/𝑝1!𝑝2!𝑝3! Number of remaining digit = 6 Thus, , n = 6 Since, three 2s & two 4s & p1 = 3, p2 = 2 The numbers of numbers beginning with 1 = 𝑛!/𝑝1!𝑝2! = 6!/(3! 2!) = (6 × 5 × 4 × 3!)/(3! × 2 × 1) = 60

Chapter 7 Class 11 Permutations and Combinations

Class 11
Important Question for exams Class 11