Last updated at May 29, 2018 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

Ex 7.1,6
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Ex 7.1,4 Important

Example 13 Important

Example 16 Important

Ex 7.3,3 Important

Ex 7.3,6 Important

Ex 7.3,10 Important

Example 19 Important

Ex 7.4, 6 Important

Ex 7.4, 8 Important

Example 23 Important You are here

Misc 3 Important

Misc 4 Important

misc 7 Important

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Misc 11 Important

Class 11

Important Question for exams Class 11

- Chapter 1 Class 11 Sets
- Chapter 2 Class 11 Relations and Functions
- Chapter 3 Class 11 Trigonometric Functions
- Chapter 4 Class 11 Mathematical Induction
- Chapter 5 Class 11 Complex Numbers
- Chapter 6 Class 11 Linear Inequalities
- Chapter 7 Class 11 Permutations and Combinations
- Chapter 8 Class 11 Binomial Theorem
- Chapter 9 Class 11 Sequences and Series
- Chapter 10 Class 11 Straight Lines
- Chapter 11 Class 11 Conic Sections
- Chapter 12 Class 11 Introduction to Three Dimensional Geometry
- Chapter 13 Class 11 Limits and Derivatives
- Chapter 14 Class 11 Mathematical Reasoning
- Chapter 15 Class 11 Statistics
- Chapter 16 Class 11 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.