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Example 23 - How many numbers greater than 1000000 can be formed - Permutation- repeating

  1. Chapter 7 Class 11 Permutations and Combinations
  2. Serial order wise
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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

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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