Last updated at March 8, 2017 by Teachoo

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

Example 1

Example 2

Example 3

Example 4

Example 5

Example 6

Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13 Important

Example 14

Example 15

Example 16 Important

Example 17

Example 18

Example 19 Important

Example 20

Example 21

Example 22

Example 23 Important You are here

Example 24

Chapter 7 Class 11 Permutations and Combinations

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .